Lecture 15. Fun
Potma, Eric
UCI Chem 1P is a preparation go General Chemistry that covers: units of measurement, dimensional analysis, significant figures; elementary concepts of volume, mass, force, pressure, energy, density, temperature, heat, work; fundamentals of atomic and molecular structure; the mole concept, stoichiometry; properties of the states of matter; gas laws; solutions concentrations. Course may be offered online. Slides:
 Okay, finished. I'm going to see valves. I'm going to do a very important chemistry. And today I will not introduce anything new. I'm probably already sure. Oh, I'm going to bring up this guy. So, very cute, very cute. I love this. Lots of parents, if they have a cool baby, they give this baby an iPad. And then they're so happy that the baby after about figures out and press something out. Technically, a little child out is an amazing thing. The fact of the matter is, who wants more babies to do? It's not really the problem. As long as you train a little puppy something eventually, it will do that. You can train a dog to pee in his place, where we can train a baby to press this button. So, in any case, these are things that are within the capabilities of us being humans. These are things that we humans can do. Now, we humans can also do things like this. Who likes to go to the causes? I like what the great amount of people likes to go to the causes. I like it too. Every night, I do want to try to go to the causes. That's very cool. I can tell you that the go to the causes are much more complex than what the calculations we do here in this class. There's no comparison. This is orders of magnitude more complex, higher order of thinking, and the kind of exercises we do in this class. Mathematically, it's just very easy to show. Even this is more difficult. Yes it is. Yes it is. So, what I'm trying to do today is try to do a couple of exercises that are within the material that we're working with, and we'll try to break it down, and we'll try to put the pieces where they belong. The mathematical steps are a multiplication or division. So we have to be able to understand the problem and just do those operations correctly. So let me first start with balancing equations. Balancing equations is like a little puzzle. It is like a little puzzle. So let me give you an example. Here's one. What do you have to do? Well, I just have to make sure that on both sides there's the same amount of atoms of each kind. I see two here. There's only one there. I see three there. And there's two there, so evidently this is not balanced. Now what happens? This is two, and this is three. So there's no way you can make this equal by just multiplying this with, let's say, two and this with two. If I multiply this with two, I generate a total of six. Six oxygens. That means I need six oxygens here too. That means if I multiply three there, I get six. And because I multiply by two, I get four of these guys, and the four of those guys. This is it. So I'm just going back and forth until everything is the same inside. Another example. Strontium phosphorus. This four here, this one there. I need three here and two there, so this evidently is not correct. I can start trying to make one of the elements correct and then balance with the other one. Now what I'm trying to do now first, let me just show you why I do this in this order. I see a E four here, and I see a one here. This one means that I can put any number in front to match the missing correlation. So this is not the one I want to balance in the beginning. I want to make sure that the phosphorus is balanced first because I see, ah, there's no coefficient there. I can put any number in front to make a match. So I start with phosphorus in this case. I have four here. I need four on the other side, so I need to multiply this by two. Two times two is four. So I generate six of these guys, the six of them there. See why I'm starting with phosphorus, right? Now let's not do that. Okay, another type of question is this. This is just kind of like a huge chemical equation not to be so scared of them. I have two Y, and this Y is an unspecified compound. It reacts with Br2, and I have two times sodium bromide. What is the missing Y? If you just have to look at this, this question, unfortunately, if you look at the equation, you just put the pieces together, really. What I see here, I have sodium and bromine. Bromine is there, sodium is not there. So I need something with sodium, and there's an iodine there too. So I need sodium iodide. Sodium iodide. If I put this one here, let's check if this is correct. Two times sodium, excuse me, two sodiums, two times iodine gives me I2. So this is a balance equation. You know, this is really actually no puzzle. This is a puzzle, I think, that falls within our union chain of deltas. There's nobody that is so challenged that they cannot solve this puzzle. The only thing is to overcome your resistance for a chemical equation. That's it. As soon as you take out the stack, then you can do the solving. Let's do this one here. Why? If you set the fill out, what is why? I see here hydrogen sulfur. Sulfur is here, hydrogen is there. And it's also oxygen is there. So oxygen is missing. Okay? So this is most likely an oxygen. So let's say this is O2. Let's see if I put O2 in there, if I actually balance the equation. Three times O2 is six octagons. Two times two is four. Two times one is two. Two plus four is six. So that means six octagons. That's right. The other one is four hydrogens, four hydrogens, and two solvers, two solvers. So this is balance, this is balance. Okay, so make sure you're just comfortable with these equations. They're little puzzles and they can be solved. All right, now let's do this. So this is a little bit of a visual illustration of the actual problem. Sometimes in the building house, if you take a question from the abstract, which is a text, into something that is actually completely an object, see the molecules and then it is built into an act. And this is the reaction that we're looking at. This should be an arrow. This is H2, and O2 is these guys. So the red ones are the oxygens, the gray ones are hydrogens, and the fourth one is water. Each water has one oxygen, two hydrogens. So the one to make a water molecule here, I need at least two of these hydrogens and one oxygen. I see one, two, three, four, five, six oxygens, so I could tentatively make six water molecules. I count six oxygens, I can make six water molecules. Okay, and that means if I can make six, I need for each water molecule two hydrogens. So each hydrogen molecule contributes two hydrogens. So if I need six water molecules, I need six of these pairs. One, two, three, four, five, six. So if I use those six plus those guys, I have these two left. So what is the situation after the reaction happens? Choose from this. I just determined that if this react reacts, I can form six water molecules out of three O2s and six H2s, and then I have two H2s left, two hydrogen molecules left. They will not react because they're running out of red spheres. So this one is not correct because I have an accessory agent which is H2. This one is also not correct because I said I could make six. This one makes three, also not correct. Look, this guy here can still react. This oxygen can react with this guy, the other oxygen with this guy can form water molecules. One, two, three, four, five, six plus two hydrogens, this must be the right one. Okay, so again, it is basically a puzzle. You just lose the idea to click, and you just have to determine how many do you form, and what do you have left and what do you have left and what. That's the problem. Okay. Now look at this. This is an interesting one. This is a sample of a compound that contains chlorine and oxygen. So this is a compound that has two elements in it, C, L, and R. Nothing else. It will react with an excess of H2. So this is the accessory agent, the other element, the other compound that's unknown at this point is the limited reagent, and they will produce two products, H, C, L, and H2O. The quantities are given in grams. So determine the empirical formula of the compound. Now, if you read this the first time, you may say, what's going on? How can I determine the empirical formula for this piece of information? That's the first response. The second response is, okay, let's first look at that. Let's see what that is. You want to visualize this problem, okay? The way I visualize it, I'm going to write a chemical equation. I'm not sure what happens with this error. Sorry, it should just be an error to the right. Apparently there's a compound that contains C, L, and O, but I don't know the substance because that's unknown. That's what I have to determine, right? So I call it X and Y. I know it reacts with H2. I also know there's a lot of H2, so H2 is not a limiting reagent. If this is a limiting reagent, okay? It produces H, C, L, and H2O. Okay, so that's that number. And I have to determine apparently what this X and Y is. How do I get that? Well, let's have a look at this problem a little closer. I see that each mole of C, L, X, or Y can produce how many HCl's? If there's X times C, L in there, if I have one mole of this, I can make X mole of HCl, right? Because each chlorine gives rise to one HCl mole. So if I have X4H in my compound, I can make XHCl's. Same for oxygen. Each water molecule has one oxygen. The oxygen must come from this. There's no other source. So if I have Y oxygen atoms, I can make one-eight quarters. What about the H? Well, I just said, no one but eight. There's enough H to take care of this. The limiting reaction is this. So the amount of HCl and H2O is determined by this compound. That means I can make X mole of HCl, Y mole of H2O. All right? Now, that is almost already the problem, because I owe us a solution to the problem, because I have in principle the quantities and moles of these guys. It's given in grams, but I know right now that I can convert grams in a flash into moles. Okay? By the time I'm done with moles. So that's what I'm going to do. I'm going to see how many moles of HCl did I generate? How many moles of H2O did I generate? The ratio between the two should be the ratio between X and Y. Okay. That's what basically what this line says. If I determine how many moles of HCl I generate, how many moles of H2O I generate, then I know the ratio between X and Y. And the ratio between X and Y is the interval. Okay? Okay, let's do this. So what I have to do is I have to calculate how many moles of this I have and how many moles of that. That is something we can do. Okay? Grams converting to moles by dividing by the mole of X. That's your conversion pattern. You don't know yet that you have to divide. You always do this conversion and check the units this has to cross out. This conversion will give you the number of moles of HCl generated. I find 6.3 or 6.4 times the number of moles of this compound. Let's assume everything happens in water. And this thing here is hydrochloric acid. How about the other guy? Okay, so this HCl corresponds to this many moles of chlorine atoms, right? Because each molecule has only one chlorine atom. So this is the amount of chlorine. All right? This is the quantity of H2O. I quickly convert that to moles by dividing by the mole of H2O. This is a division. What I get is 2.24 times 10 to the minus 2 moles of H2O. Each H2O molecule contributes one oxygen. That means I have 2.24 times 10 to the minus 2 moles of oxygen. So this is then X moles of this, Y moles of that. So this is X is Y. I have to calculate the ratio between the two. So that means I can find the empirical ratio between chlorine and oxygen. Chlorine I have the quantity of, oxygen I have the quantity of. The ratio between the two is the ratio between the subscripts. How many moles of C I have relative to the amount of moles of oxygen? So if I divide one by the other, let's just put oxygen on top. It doesn't have to be a quantity, right? Just choose one. Oxygen on top, divide that by the amount of moles of chlorine. The ratio is 3.5. That means for each chlorine I have 3.5 oxygens. Right? For each chlorine I have 3.5 oxygens. That's what this calculation says. We also know that in the empirical formula, we don't have fractions. It should be an integer. So it's not close to an integer number. Is that a problem? Not necessarily. I could just multiply it with a number so that I can make an integer. So let's look at that. I have here for each one Cl, 3.5 oxygens. That's what I just determined. For each one Cl, for each one mole of Cl, I have 3.5 moles of oxygen. If I multiply this whole unit by two, I get this. Cl2 of 7. That is the empirical formula. It's not the molecular formula yet. That's not a question. The question is determining the empirical formula. And the answer is right here. So this is a slightly different question. But again, with the same kind of thinking. How many atoms do I have? How many moles do I have? And how do they relate to one another? If I have this much of this, how much do I have to be audible? That is the kind of thinking I have for applying the multiply. I have a question. Why do you multiply it by two? The question is why do you multiply it by two? Because an empirical formula should give you the substance in integers. This is not an integer. This is 3.5. You have to do that. You have to multiply it with the lowest multiplier that gives you the degree. And that is what we do here. The lowest multiplier is two because two times 3.5 is seven. That is an integer. And in the process, I have two here as well. The empirical formula CO2, 07. Yes, another question. I have a question. I said hydrogen is an excess. I did not have to worry about the high. But for each molecule of HCl, I need one atom of H. One hydrogen. For each molecule of oxygen of water, I need two atoms of hydrogen. So it depends on if you look at this guy or this guy. So there are different ratios. You need more H than you need oxygen and CO. But fortunately, there is an abundance of Hs in the process. So I do not have to worry about it. The pressure is not about that. The pressure is typically about that. And then we just look at how many Hs in that. It needs to be tonic energy. And how much we need. And how much H we have consumed. You can do that calculation as well. Okay. So let's do the following. Another reaction. Remember, these typos are arrows. C6 H14 is converted into C2 H4. That is a reaction. There is also other stuff that I do not have to worry about. Basically, this means that I have one mole of this that is converted into one mole of that. The ratio is one to one. If the yield of this process is 42.5%, what mass of this starting material, which is called hexane, must be there to produce 481 grams of the product, which is called ethylene. So, in other words, your boss tells you, I need from you 481 grams of ethylene. Okay? And he also tells you what. This reaction is on very efficient. Efficiency is exactly 42.1, 42.5. And I want you to determine how much hexane do I need to pour. Can you give me that answer finally? He said, ah, I can do that. Come on, I need only three. All right. So what you do is this. Okay? You would like to know how many moles of the product do I need? How many moles of the product do I need? Again, grams, you want to move very quickly, convert grams all the way to moles. So, something like a countable number, which is moles. 48, 481 grams divided by the molar mass of ethylene. Okay? Which is 28.05. Says 70.1 moles of ethylene. The boss apparently needs 70.1 moles of ethylene. That's what he needs. So that means if you have 70.1 moles of ethylene, how much starting material do we need? Well, the reaction says it reacts one to one. One mole of hexane, the starting material, the reagent produces one mole of this. Okay? So, if this reaction were to be one hundred percent efficient, you would need 70.1 moles of the starting material to react one to one. 70.1 moles in, 70.1 moles out. Okay? However, the efficiency is 42.5. So, you have to take that into consideration. Apparently, you need more than 70.1 because efficiency is rather large. So, how do you do that? Okay? So, this is what I just said. So, take into account the yield. The yield expressed in a fraction times the amount that you need will be 70.1 moles. Okay? You need 70.1 moles. That should be the outcome. Okay? You need 70.1 moles. But you need to put more in because you multiply something as small as one. Okay? So, the fraction corresponding to this percentage is 0.425 times the amount that you have to put in should give you the amount that you have to give to your supervisor. So, that means the amount that I need, I can quickly calculate, 17.1 moles divided by this fraction 40.2 moles is the amount you have to start with. Now, this is the amount of moles. The question also says, I need it in grams. So, what you have to do is, you have to multiply this by the molar mass to get the amount of grams in grams. Is this set clear? Does that make sense? Alright. So, the multiplication of the molar mass you can do for yourself when you have the answer. So, we just move on to this. This question. This question is the same kind of question, but the phrasing is a little different. Now, don't be shaken by this. It is again the same kind of thinking. And the reason why I like this example is it forces you to really think about these things as countable objects. Some of this, some of that, the reactometry and so forth. Okay, so here's another reaction. Methane reacts with oxygen, combustion reaction with methane. CO2 and 2H2O comes out. I have to determine the number of excess reagent. So, excess reagent means what is left in the pot after this reaction is over. Okay? How many do I have in the beginning? Well, it says I have 70 molecules of methane and 142 of oxygen. The funny thing about this question is it doesn't say how many moles you have or how many grams you have. Okay? That is another problem. That's the same thing. Another molecule suggests another molecule. But it is specified in an actual number for a mole that doesn't matter. Irrigate with the numbers. Okay, so what I have to do is to determine which one of these reagents is the limited area. In order to do that, I play the same trick as before. I want to know what is the mole ratio between these two guys according to the equation, the balanced equation. It says you need two moles of oxygen for each one mole of methane. And that means the ratio of 0.5. If you put methane on top, the ratio will be 0.5. The next thing I'm going to do, so this is the mole ratio. Mole ratio. I want to determine the mole ratio that I actually have. Again, the number of moles is not given. The number of molecules is given. But what is a mole? A mole is a number. It is 6.05. So, we're at 6.02, the time is 6.33. So, I think you can convert this to moles if you want. You multiply this by moles, by a mole number, this by a mole number. But the ratio will change. Because if you multiply this by a mole, this by a mole, you divide them out, you just see what I'm saying. The ratio is determined by the proportionality between the two. Whether you have a mole of it, or just a countable number of molecules on your fingers, the ratio is still the same. So, I can determine the mole ratio simply by dividing the quantities of these numbers. Of the number of molecules that I have. Okay? So, let's just move on with this. So, 70 molecules of CH4, 142 molecules of O2. So, this is the actual mole ratio. Not given in moles, but in just molecules. Because the ratio between molecules and the ratio between moles is the same. Okay? I can convert this to moles if you want. It doesn't matter, because you convert also this one to more. The ratio between the two is 0.49. 0.49 is a little bit smaller than 0.5. I think the numerator is the limiting reagent. I eat the molecule of one dot first. You will burn through the 70 sooner than you burn through the 140. Okay? So, it's a smaller than 0.5. That means CH4, methane, is the limiting reagent. That means the other guy is the excess reagent. Okay? This O2. That means I have something left in the pot after the reaction has completed. Okay? So, how many molecules of oxygen did react? Same kind of calculation. 70 molecules of methane were there. That's the limiting reagent. I will convert the number of reactive methane to the number of reactive oxygen through the mole ratio. Okay? Two molecules of oxygen for each one molecule of CH4. That's what the equation says. That's the ratio between the two. For each one molecule of methane, any two molecules of oxygen. So, each one molecule of methane, two molecules of oxygen. This one goes at the bottom because I want to cross this out. I want to go to the number of molecules of oxygen that has reacted. 70 times 2 equals 140 molecules of oxygen have reacted in this process. Okay? So, how many do I have left? Yes, you have 142 minus 140, two molecules of oxygen. Now, mind you, this is exactly what we did before. Previously, we did it at quantities expressed in moles. Now, we do it in the physical number of molecules. But it's exactly the same because you work with ratios. You work with ratios. So, you're going to replicate them exactly the same. I'm going to do another one which is virtually the same. Just to get used to this. The same kind of question, a different reaction. This is also open-bush reaction. It's a different kind of molecule. But the derivative of this should be the same. Okay? Calculate the number of excess reagent molecules after this reaction has proceeded. I'll do the same thing again. First, the mole ratio according to the equation. The equation says two moles of this reagent for each 15 moles of oxygen that ratio is 21.36. That's what I call the Bellos ratio. The ratio stipulated by the Bellos chemical equation. Then, I want to see what I actually have. What is the ratio of the chemicals to reagents that I actually have? Well, here are the numbers that I actually have. Okay? The same kind of calculation. 74 divided by 585 equals 0.126. This number? Less than that. That means the numerator is going to be my limiting reagent. Okay? So, C5H10 is the limiting reagent. You can always determine how many molecules have reacted. All the limiting reagent react, and when the limiting reagent reacts, it takes with it, in this ratio, the oxygen. That means it's compared by how many oxygen's have reacted in the process. So, that conversion is, again, number of molecules of limiting reagent is reacting. It takes with it, for each two of this, 15 molecules of oxygen. That's the mole ratio of the two molecules. Okay? So, 74 molecules of this reagent react in this ratio. Each two molecules reacting take with it 15 molecules of oxygen. That means 55 oxygen molecules have reacted. The conversion here is from number of molecules of C5H10 to number of molecules of O2. O2 goes on top. This guy's going at the bottom. This is what I'm always doing. We need to convert from one quality into the next. How many did I have in the beginning? 585. This many have reacted. So, 555 minus this 30 molecules of oxygen are left. They cannot react. Why? Because there's no other reagent anymore. It can't go on. Okay. I think we've got to run for it. I do one more of this, and then we'll do a final thing accepting SSI. Okay. So, we do it all quick. Another reagent is safe. Unlike repetition, it really gets you into the mood. And these, like I said, dissolve white foam. Okay. So, another compound reacting with an oxygen oxide producing these two compounds. And again, I've got the excess reagent. So, he is the truncable limited reagent. What you do is look at the mole ratio before we get up to the chemical equation. Two to ten, at least, points. You compare that with the actual number that you have. Okay. That is, this number, compared to that number. Whether the number is stipulated or given in grams or moles or number of molecules, it doesn't matter. It can give you a ratio. Okay. Grams always have to be converted to moles first. And then you get the mole ratio. And the number of molecules that have a number of molecules is also mole ratio. This number divided by that makes you 0.182. Sorry, 181. That is less than 0.2. That means numerator, again, is the limited reagent. You have all the numerators here. So, there are numerators always the limited reagent. You could have as well been the denominator. In other words, it just happens to be the numerator. Okay. So, the numerator is the limited reagent. In this case, this is a di-I-D pentoxide. That means I take the number of limited reagent, okay, and I convert that into the number of excess reagent. Reactive. How? Multiply that in a ratio. So, I take the number of the three units of this, I take with it 10 units of nitrogen oxide. This times that gives you the number of nitrogen oxide mole to mole with half the reactant in the process, 980. This is the amount that I had in the beginning. So, I take that, subtract this from it. I have exactly 100 units left. Units is nothing but the mole. Okay. So, this is the flow of these kind of aeroexercises. You can see that, and hopefully that will not be too hard for you. Okay. So, let's do a couple of minutes. Let's say there's no life that we have. We apply to an actual problem. So, the problem of exploring cars. So, we've seen it in some of the movies. We've seen it in some more of the left and right. The same question we've been trying to solve before the problem. It seems like it is extremely easy to realize. Take a car and explore it. Let's see if this really is that easy. Okay. What happens in this process is there's a chemical reaction. One is exploding, gas will leave. What does gas leave? Mostly, captain. That's captain. And it reacts with oxygen. That's the arrow that should go to the right. And it generates C2O2 and H2O2. This is combustion reaction. And it's boundless being all red. So, when does this explosion happen? It happens when this, the meeting of O2, but half day, is very efficient. So, if these chemicals are present in the right proportion, they can very actively react and explore. If there's not enough oxygen or not enough captain, then the chemicals have to wait till they leave and it slows down the process of the laserous process. So, what I really want is I want this reaction to be complete. I want the reagents to be present in the right proportions and, in the optimum, react with conditions. That's a long way to say that what I need is, for each one mole of heptane, I need 11 moles of oxygen. That's what I want in my tank. So, let's look at that. Here's a tank. A typical tank is about 14 gallons, converted immediately to metric units. It's about 53 liters. This is a very rough definition. Let's make it a 9-number, 53 liters is the approximate volume of a typical gas tank. The density of heptane will need that, this number right here, given in grams per mil. And the density of oxygen is also given, remember oxygen is a gas, okay, so that is 22.4 liters per mole. And here's the reaction once again. So, what do I need? Well, I just said it. What you want to do is you want to have quantities of oxygen and heptane that are correct for reacting explosively. That means 11 moles of oxygen for each one mole of heptane. So, the ratio needs to be 11. Okay, now let's see intuitively, we think this is what your first gas would be. I need as much heptane as possible to invest, explosion time, completely fill the tank. So, let's do that. Let's assume the tank is a 24, because I think in that case, more heptane means bigger explosion. So, 53 liters, that's the whole thing, tank, it's all full of heptane now. How many moles of heptane is that? That's a couple, I think. So, the gas times the density converts a volume into a mass. So, I multiply this by this number here. This is the density, 0.684 grams per milliliter or kilograms per liter. Remember, grams per milliliter is the same unit as kilograms per liters. It has the same value of the density. The value of the density is the same, it's expressed in kilograms per liter, more grams per milliliter. Or you can use that, and that's helpful because this is liter times liter, so this crosses out, now you have the mass of heptane in the tank. Okay, if the mass is expressed in kilograms, I can express mass in moles by the value of the volume, mole mass. So, watch out if you can do that correctly. Usually, the mole mass is given in grams per mole, but now I'm going to do kilograms per mole. So, that means you divide it by a thousand. The kilogram across out, what you get is moles of heptane. Same calculation that we did. You have the mass, you divide it by the mole mass. The mass is known as kilograms divided by the mole mass, and you can express in kilograms per mole. Heptane is about 100 grams per mole, that means 0.1 kilograms per mole. I find 3.6 times 10 to the second mole of heptane. So, this is the amount of heptane that I got. Okay, so I need relative to heptane, 11 times more oxygen for this we have to have. Well, the problem is, I don't have any oxygen because the thing is full. There is no oxygen to react with. The thing is completely full, there is really no oxygen that can officially get to the heptane. It's not in the same place. This is a Q student pipe, that's R-I. There's no oxygen there for all the molecules of heptane to react with. So, this is absolutely not the situation that we're dealing with. If you're dancing to school, you're very sick. You're a lot of swell. You can fire for yourself, you can make a car of fire, you can make a car of speech, you can see, boom, swell. Make sure you jump out before the falls. So, what we need, what we need is we need more oxygen. Okay, we need more oxygen to live here because the reaction needs oxygen, there's no oxygen in the present yet, so I need to get some heptane out and put some oxygen in. Okay, so let's assume for the following, let's say I have only one liter of heptane there for me, and let's assume for all kinds of purposes that the air is 100% hydrogen, which is not sure. Well, that's just a super-design. One liter of heptane in tank, one liter, it'll be the same calculation, let's see how many moles are there. One liter times the density gives me mass, mass divided by mole, mass gives me moles. I complete this calculation, I find 6.8 moles of heptane, not that the quantity of heptane. And now I need to see how many moles are hydrogen. So, let's see, if this is one mole of heptane, the total of 53, so that means 52 liters is hydrogen, and I assume now, again, all the air is hydrogen. It's not sure what it's going to say. So, let's just do that. So, what I do now, I convert liters directly to mole, because I have this conversion here, 22.4 liters per mole. So, if I multiply 52 liters by this conversion factor, I put liters at the bottom and mole on top, I directly convert the mole. I find 2.3 moles of hydrogen. So, 52 liters of hydrogen gas, apparently, is 2.3 moles. Well, due to the division in 2.3 moles by 6.8 moles of heptane, it's 0.33, I need 11. This is W plus. So, even in a situation where 100% oxygen in the air, one liter of heptane is maintained, and it won't explode. Very very unlikely. So, when it actually does it, I want this thing to explode. I'm not giving up. So, I'm going to take almost all the heptane out of the tank. Let's just say that I, you know, I think it's almost completely filled with oxygen. That means 53 liters is now oxygen, and I'm going to convert how many moles that is to the same calculation. 53, this conversion factor from up here, this one at the bottom, this on top, gives me 2.4 moles of oxygen. So, I have 2.4 moles of oxygen in my tank. How much heptane do I need to make this fully react? Again, under the assumption, there's 100% oxygen in the air. Okay, well, the ratio needs to be 1 to 11. 1 to 11. What we need is 2.4 moles of oxygen. I take the ratio of oxygen at the bottom, 11 moles of oxygen for each one mole of heptane. So, effectively, you divide this thing by 11. Okay? This here is the number of moles of heptane. Okay? I multiply that by the molar mass. That means how many grams I have. So, this is the mass of heptane. Okay? And then I'm going to multiply that by, let's see, by this density conversion. So, I convert the mass into volume. So, I divide it by the mass. So, I'm going to do a free conversion here. First, from number of oxygen to number of heptane that I need, moles of heptane converted into grams of heptane, and then mass, grams, converted into volume. Through the density, I divide that by this. You see, grams at the bottom across out, I need a time to answer in milliliters, 32 milliliters of heptane. 32 milliliters is about a quarter of a total. Okay? 1-eighth on a cup. 1-eighth on a cup is what you need approximately to have 2.4 moles of oxygen to react explosively with heptane. Okay? So, that is basically what it says. You need a small fill of gasoline on your tank to make that thing explode. And then you also need about 300 percent oxygen in here. Then you can have a very, very strong flow of oxygen. So, less is better, less is better. Alright, so this is an example of calculations. It uses mole ratios. It also uses things like this, volumes, densities, right? These are things we have seen in the first part of this class. We're getting close to the first couple of lectures. We're all about densities, volumes, so forth. This is a real-life situation. You could do these calculations with the tools that we've talked about. One more question. Yes? Yes, hold on a second. There's a nice question here. Okay? So, does that necessarily mean that with, you know, the degree of the particles on my tank, it's really more density? The question is, why does it explode in a movie? Well, first of all, in a movie, it's for the good sticks of dynamite, to make it explode, right? But particles can explode. And typically, it comes to explode when you have a leak, and stuff drips from the floor. It's a small film with what you did, and that's what it's set up. It typically in itself, it won't have conditions or something. Okay, whoa!
 Okay, finished.  I'm going to see valves.  I'm going to do a very important chemistry.  And today I will not introduce anything new.  I'm probably already sure.  Oh, I'm going to bring up this guy.  So, very cute, very cute.  I love this.  Lots of parents, if they have a cool baby,  they give this baby an iPad.  And then they're so happy that the baby after about figures out  and press something out.  Technically, a little child out is an amazing thing.  The fact of the matter is, who wants more babies to do?  It's not really the problem.  As long as you train a little puppy something eventually,  it will do that.  You can train a dog to pee in his place,  where we can train a baby to press this button.  So, in any case, these are things that are within the capabilities  of us being humans.  These are things that we humans can do.  Now, we humans can also do things like this.  Who likes to go to the causes?  I like what the great amount of people likes to go to the causes.  I like it too.  Every night, I do want to try to go to the causes.  That's very cool.  I can tell you that the go to the causes are much more complex  than what the calculations we do here in this class.  There's no comparison.  This is orders of magnitude more complex,  higher order of thinking,  and the kind of exercises we do in this class.  Mathematically, it's just very easy to show.  Even this is more difficult.  Yes it is.  Yes it is.  So, what I'm trying to do today is try to do a couple of exercises  that are within the material that we're working with,  and we'll try to break it down,  and we'll try to put the pieces where they belong.  The mathematical steps are a multiplication or division.  So we have to be able to understand the problem  and just do those operations correctly.  So let me first start with balancing equations.  Balancing equations is like a little puzzle.  It is like a little puzzle.  So let me give you an example.  Here's one.  What do you have to do?  Well, I just have to make sure that on both sides  there's the same amount of atoms of each kind.  I see two here.  There's only one there.  I see three there.  And there's two there, so evidently this is not balanced.  Now what happens?  This is two, and this is three.  So there's no way you can make this equal by just multiplying this  with, let's say, two and this with two.  If I multiply this with two, I generate a total of six.  Six oxygens.  That means I need six oxygens here too.  That means if I multiply three there, I get six.  And because I multiply by two, I get four of these guys,  and the four of those guys.  This is it.  So I'm just going back and forth  until everything is the same inside.  Another example.  Strontium phosphorus.  This four here, this one there.  I need three here and two there,  so this evidently is not correct.  I can start trying to make one of the elements correct  and then balance with the other one.  Now what I'm trying to do now first,  let me just show you why I do this in this order.  I see a E four here, and I see a one here.  This one means that I can put any number in front  to match the missing correlation.  So this is not the one I want to balance in the beginning.  I want to make sure that the phosphorus is balanced first  because I see, ah, there's no coefficient there.  I can put any number in front to make a match.  So I start with phosphorus in this case.  I have four here.  I need four on the other side, so I need to multiply this by two.  Two times two is four.  So I generate six of these guys, the six of them there.  See why I'm starting with phosphorus, right?  Now let's not do that.  Okay, another type of question is this.  This is just kind of like a huge chemical equation  not to be so scared of them.  I have two Y, and this Y is an unspecified compound.  It reacts with Br2,  and I have two times sodium bromide.  What is the missing Y?  If you just have to look at this, this question, unfortunately,  if you look at the equation, you just put the pieces together, really.  What I see here, I have sodium and bromine.  Bromine is there, sodium is not there.  So I need something with sodium, and there's an iodine there too.  So I need sodium iodide.  Sodium iodide.  If I put this one here, let's check if this is correct.  Two times sodium, excuse me, two sodiums,  two times iodine gives me I2.  So this is a balance equation.  You know, this is really actually no puzzle.  This is a puzzle, I think, that falls within our union chain of deltas.  There's nobody that is so challenged that they cannot solve this puzzle.  The only thing is to overcome your resistance for a chemical equation.  That's it.  As soon as you take out the stack, then you can do the solving.  Let's do this one here.  Why?  If you set the fill out, what is why?  I see here hydrogen sulfur.  Sulfur is here, hydrogen is there.  And it's also oxygen is there.  So oxygen is missing.  Okay?  So this is most likely an oxygen.  So let's say this is O2.  Let's see if I put O2 in there, if I actually balance the equation.  Three times O2 is six octagons.  Two times two is four.  Two times one is two.  Two plus four is six.  So that means six octagons.  That's right.  The other one is four hydrogens, four hydrogens, and two solvers, two solvers.  So this is balance, this is balance.  Okay, so make sure you're just comfortable with these equations.  They're little puzzles and they can be solved.  All right, now let's do this.  So this is a little bit of a visual illustration of the actual problem.  Sometimes in the building house, if you take a question from the abstract, which is a text,  into something that is actually completely an object,  see the molecules and then it is built into an act.  And this is the reaction that we're looking at.  This should be an arrow.  This is H2, and O2 is these guys.  So the red ones are the oxygens, the gray ones are hydrogens, and the fourth one is water.  Each water has one oxygen, two hydrogens.  So the one to make a water molecule here, I need at least two of these hydrogens and one oxygen.  I see one, two, three, four, five, six oxygens, so I could tentatively make six water molecules.  I count six oxygens, I can make six water molecules.  Okay, and that means if I can make six, I need for each water molecule two hydrogens.  So each hydrogen molecule contributes two hydrogens.  So if I need six water molecules, I need six of these pairs.  One, two, three, four, five, six.  So if I use those six plus those guys, I have these two left.  So what is the situation after the reaction happens?  Choose from this.  I just determined that if this react reacts, I can form six water molecules out of three O2s and six H2s,  and then I have two H2s left, two hydrogen molecules left.  They will not react because they're running out of red spheres.  So this one is not correct because I have an accessory agent which is H2.  This one is also not correct because I said I could make six.  This one makes three, also not correct.  Look, this guy here can still react.  This oxygen can react with this guy, the other oxygen with this guy can form water molecules.  One, two, three, four, five, six plus two hydrogens, this must be the right one.  Okay, so again, it is basically a puzzle.  You just lose the idea to click, and you just have to determine how many do you form,  and what do you have left and what do you have left and what.  That's the problem.  Okay.  Now look at this.  This is an interesting one.  This is a sample of a compound that contains chlorine and oxygen.  So this is a compound that has two elements in it, C, L, and R. Nothing else.  It will react with an excess of H2.  So this is the accessory agent, the other element, the other compound that's unknown at this point  is the limited reagent, and they will produce two products, H, C, L, and H2O.  The quantities are given in grams.  So determine the empirical formula of the compound.  Now, if you read this the first time, you may say, what's going on?  How can I determine the empirical formula for this piece of information?  That's the first response.  The second response is, okay, let's first look at that.  Let's see what that is.  You want to visualize this problem, okay?  The way I visualize it, I'm going to write a chemical equation.  I'm not sure what happens with this error.  Sorry, it should just be an error to the right.  Apparently there's a compound that contains C, L, and O,  but I don't know the substance because that's unknown.  That's what I have to determine, right?  So I call it X and Y.  I know it reacts with H2.  I also know there's a lot of H2, so H2 is not a limiting reagent.  If this is a limiting reagent, okay?  It produces H, C, L, and H2O.  Okay, so that's that number.  And I have to determine apparently what this X and Y is.  How do I get that?  Well, let's have a look at this problem a little closer.  I see that each mole of C, L, X, or Y can produce how many HCl's?  If there's X times C, L in there,  if I have one mole of this, I can make X mole of HCl, right?  Because each chlorine gives rise to one HCl mole.  So if I have X4H in my compound, I can make XHCl's.  Same for oxygen.  Each water molecule has one oxygen.  The oxygen must come from this.  There's no other source.  So if I have Y oxygen atoms, I can make one-eight quarters.  What about the H?  Well, I just said, no one but eight.  There's enough H to take care of this.  The limiting reaction is this.  So the amount of HCl and H2O is determined by this compound.  That means I can make X mole of HCl, Y mole of H2O.  All right?  Now, that is almost already the problem,  because I owe us a solution to the problem,  because I have in principle the quantities and moles of these guys.  It's given in grams, but I know right now that I can convert grams  in a flash into moles.  Okay?  By the time I'm done with moles.  So that's what I'm going to do.  I'm going to see how many moles of HCl did I generate?  How many moles of H2O did I generate?  The ratio between the two should be the ratio between X and Y.  Okay.  That's what basically what this line says.  If I determine how many moles of HCl I generate,  how many moles of H2O I generate,  then I know the ratio between X and Y.  And the ratio between X and Y is the interval.  Okay?  Okay, let's do this.  So what I have to do is I have to calculate how many moles of this I have  and how many moles of that.  That is something we can do.  Okay?  Grams converting to moles by dividing by the mole of X.  That's your conversion pattern.  You don't know yet that you have to divide.  You always do this conversion and check the units this has to cross out.  This conversion will give you the number of moles of HCl generated.  I find 6.3 or 6.4 times the number of moles of this compound.  Let's assume everything happens in water.  And this thing here is hydrochloric acid.  How about the other guy?  Okay, so this HCl corresponds to this many moles of chlorine atoms, right?  Because each molecule has only one chlorine atom.  So this is the amount of chlorine.  All right?  This is the quantity of H2O.  I quickly convert that to moles by dividing by the mole of H2O.  This is a division.  What I get is 2.24 times 10 to the minus 2 moles of H2O.  Each H2O molecule contributes one oxygen.  That means I have 2.24 times 10 to the minus 2 moles of oxygen.  So this is then X moles of this, Y moles of that.  So this is X is Y.  I have to calculate the ratio between the two.  So that means I can find the empirical ratio between chlorine and oxygen.  Chlorine I have the quantity of, oxygen I have the quantity of.  The ratio between the two is the ratio between the subscripts.  How many moles of C I have relative to the amount of moles of oxygen?  So if I divide one by the other, let's just put oxygen on top.  It doesn't have to be a quantity, right?  Just choose one.  Oxygen on top, divide that by the amount of moles of chlorine.  The ratio is 3.5.  That means for each chlorine I have 3.5 oxygens.  Right?  For each chlorine I have 3.5 oxygens.  That's what this calculation says.  We also know that in the empirical formula,  we don't have fractions.  It should be an integer.  So it's not close to an integer number.  Is that a problem?  Not necessarily.  I could just multiply it with a number so that I can make an integer.  So let's look at that.  I have here for each one Cl, 3.5 oxygens.  That's what I just determined.  For each one Cl, for each one mole of Cl,  I have 3.5 moles of oxygen.  If I multiply this whole unit by two, I get this.  Cl2 of 7.  That is the empirical formula.  It's not the molecular formula yet.  That's not a question.  The question is determining the empirical formula.  And the answer is right here.  So this is a slightly different question.  But again, with the same kind of thinking.  How many atoms do I have?  How many moles do I have?  And how do they relate to one another?  If I have this much of this, how much do I have to be audible?  That is the kind of thinking I have for applying the multiply.  I have a question.  Why do you multiply it by two?  The question is why do you multiply it by two?  Because an empirical formula should give you the substance in integers.  This is not an integer.  This is 3.5.  You have to do that.  You have to multiply it with the lowest multiplier that gives you the degree.  And that is what we do here.  The lowest multiplier is two because two times 3.5 is seven.  That is an integer.  And in the process, I have two here as well.  The empirical formula CO2, 07.  Yes, another question.  I have a question.  I said hydrogen is an excess.  I did not have to worry about the high.  But for each molecule of HCl, I need one atom of H.  One hydrogen.  For each molecule of oxygen of water, I need two atoms of hydrogen.  So it depends on if you look at this guy or this guy.  So there are different ratios.  You need more H than you need oxygen and CO.  But fortunately, there is an abundance of Hs in the process.  So I do not have to worry about it.  The pressure is not about that.  The pressure is typically about that.  And then we just look at how many Hs in that.  It needs to be tonic energy.  And how much we need.  And how much H we have consumed.  You can do that calculation as well.  Okay.  So let's do the following.  Another reaction.  Remember, these typos are arrows.  C6 H14 is converted into C2 H4.  That is a reaction.  There is also other stuff that I do not have to worry about.  Basically, this means that I have one mole of this that is converted into one mole of that.  The ratio is one to one.  If the yield of this process is 42.5%, what mass of this starting material, which is called hexane,  must be there to produce 481 grams of the product, which is called ethylene.  So, in other words, your boss tells you, I need from you 481 grams of ethylene.  Okay?  And he also tells you what.  This reaction is on very efficient.  Efficiency is exactly 42.1, 42.5.  And I want you to determine how much hexane do I need to pour.  Can you give me that answer finally?  He said, ah, I can do that.  Come on, I need only three.  All right.  So what you do is this.  Okay?  You would like to know how many moles of the product do I need?  How many moles of the product do I need?  Again, grams, you want to move very quickly, convert grams all the way to moles.  So, something like a countable number, which is moles.  48, 481 grams divided by the molar mass of ethylene.  Okay?  Which is 28.05.  Says 70.1 moles of ethylene.  The boss apparently needs 70.1 moles of ethylene.  That's what he needs.  So that means if you have 70.1 moles of ethylene, how much starting material do we need?  Well, the reaction says it reacts one to one.  One mole of hexane, the starting material, the reagent produces one mole of this.  Okay?  So, if this reaction were to be one hundred percent efficient,  you would need 70.1 moles of the starting material to react one to one.  70.1 moles in, 70.1 moles out.  Okay?  However, the efficiency is 42.5.  So, you have to take that into consideration.  Apparently, you need more than 70.1 because efficiency is rather large.  So, how do you do that?  Okay?  So, this is what I just said.  So, take into account the yield.  The yield expressed in a fraction times the amount that you need will be 70.1 moles.  Okay?  You need 70.1 moles.  That should be the outcome.  Okay?  You need 70.1 moles.  But you need to put more in because you multiply something as small as one.  Okay?  So, the fraction corresponding to this percentage is 0.425 times the amount  that you have to put in should give you the amount that you have to give to your supervisor.  So, that means the amount that I need, I can quickly calculate,  17.1 moles divided by this fraction 40.2 moles is the amount you have to start with.  Now, this is the amount of moles.  The question also says, I need it in grams.  So, what you have to do is, you have to multiply this by the molar mass  to get the amount of grams in grams.  Is this set clear?  Does that make sense?  Alright.  So, the multiplication of the molar mass you can do for yourself when you have the answer.  So, we just move on to this.  This question.  This question is the same kind of question, but the phrasing is a little different.  Now, don't be shaken by this.  It is again the same kind of thinking.  And the reason why I like this example is it forces you to really think about these things as  countable objects.  Some of this, some of that, the reactometry and so forth.  Okay, so here's another reaction.  Methane reacts with oxygen, combustion reaction with methane.  CO2 and 2H2O comes out.  I have to determine the number of excess reagent.  So, excess reagent means what is left in the pot after this reaction is over.  Okay?  How many do I have in the beginning?  Well, it says I have 70 molecules of methane and 142 of oxygen.  The funny thing about this question is it doesn't say how many moles you have  or how many grams you have.  Okay?  That is another problem.  That's the same thing.  Another molecule suggests another molecule.  But it is specified in an actual number for a mole that doesn't matter.  Irrigate with the numbers.  Okay, so what I have to do is to determine which one of these reagents  is the limited area.  In order to do that, I play the same trick as before.  I want to know what is the mole ratio between these two guys  according to the equation, the balanced equation.  It says you need two moles of oxygen for each one mole of methane.  And that means the ratio of 0.5.  If you put methane on top, the ratio will be 0.5.  The next thing I'm going to do, so this is the mole ratio.  Mole ratio.  I want to determine the mole ratio that I actually have.  Again, the number of moles is not given.  The number of molecules is given.  But what is a mole?  A mole is a number.  It is 6.05.  So, we're at 6.02, the time is 6.33.  So, I think you can convert this to moles if you want.  You multiply this by moles, by a mole number, this by a mole number.  But the ratio will change.  Because if you multiply this by a mole, this by a mole,  you divide them out, you just see what I'm saying.  The ratio is determined by the proportionality between the two.  Whether you have a mole of it, or just a countable number of molecules on your fingers,  the ratio is still the same.  So, I can determine the mole ratio simply by dividing the quantities of these numbers.  Of the number of molecules that I have.  Okay?  So, let's just move on with this.  So, 70 molecules of CH4, 142 molecules of O2.  So, this is the actual mole ratio.  Not given in moles, but in just molecules.  Because the ratio between molecules and the ratio between moles is the same.  Okay?  I can convert this to moles if you want.  It doesn't matter, because you convert also this one to more.  The ratio between the two is 0.49.  0.49 is a little bit smaller than 0.5.  I think the numerator is the limiting reagent.  I eat the molecule of one dot first.  You will burn through the 70 sooner than you burn through the 140.  Okay?  So, it's a smaller than 0.5.  That means CH4, methane, is the limiting reagent.  That means the other guy is the excess reagent.  Okay?  This O2.  That means I have something left in the pot after the reaction has completed.  Okay?  So, how many molecules of oxygen did react?  Same kind of calculation.  70 molecules of methane were there.  That's the limiting reagent.  I will convert the number of reactive methane to the number of reactive oxygen through the mole ratio.  Okay?  Two molecules of oxygen for each one molecule of CH4.  That's what the equation says.  That's the ratio between the two.  For each one molecule of methane, any two molecules of oxygen.  So, each one molecule of methane, two molecules of oxygen.  This one goes at the bottom because I want to cross this out.  I want to go to the number of molecules of oxygen that has reacted.  70 times 2 equals 140 molecules of oxygen have reacted in this process.  Okay?  So, how many do I have left?  Yes, you have 142 minus 140, two molecules of oxygen.  Now, mind you, this is exactly what we did before.  Previously, we did it at quantities expressed in moles.  Now, we do it in the physical number of molecules.  But it's exactly the same because you work with ratios.  You work with ratios.  So, you're going to replicate them exactly the same.  I'm going to do another one which is virtually the same.  Just to get used to this.  The same kind of question, a different reaction.  This is also open-bush reaction.  It's a different kind of molecule.  But the derivative of this should be the same.  Okay?  Calculate the number of excess reagent molecules after this reaction has proceeded.  I'll do the same thing again.  First, the mole ratio according to the equation.  The equation says two moles of this reagent for each 15 moles of oxygen that ratio is 21.36.  That's what I call the Bellos ratio.  The ratio stipulated by the Bellos chemical equation.  Then, I want to see what I actually have.  What is the ratio of the chemicals to reagents that I actually have?  Well, here are the numbers that I actually have.  Okay? The same kind of calculation.  74 divided by 585 equals 0.126.  This number?  Less than that.  That means the numerator is going to be my limiting reagent.  Okay?  So, C5H10 is the limiting reagent.  You can always determine how many molecules have reacted.  All the limiting reagent react, and when the limiting reagent reacts,  it takes with it, in this ratio, the oxygen.  That means it's compared by how many oxygen's have reacted in the process.  So, that conversion is, again,  number of molecules of limiting reagent is reacting.  It takes with it, for each two of this, 15 molecules of oxygen.  That's the mole ratio of the two molecules.  Okay?  So, 74 molecules of this reagent react in this ratio.  Each two molecules reacting take with it 15 molecules of oxygen.  That means 55 oxygen molecules have reacted.  The conversion here is from number of molecules of C5H10 to number of molecules of O2.  O2 goes on top.  This guy's going at the bottom.  This is what I'm always doing.  We need to convert from one quality into the next.  How many did I have in the beginning?  585.  This many have reacted.  So, 555 minus this 30 molecules of oxygen are left.  They cannot react.  Why?  Because there's no other reagent anymore.  It can't go on.  Okay.  I think we've got to run for it.  I do one more of this, and then we'll do a final thing accepting SSI.  Okay.  So, we do it all quick.  Another reagent is safe.  Unlike repetition, it really gets you into the mood.  And these, like I said, dissolve white foam.  Okay.  So, another compound reacting with an oxygen oxide producing these two compounds.  And again, I've got the excess reagent.  So, he is the truncable limited reagent.  What you do is look at the mole ratio before we get up to the chemical equation.  Two to ten, at least, points.  You compare that with the actual number that you have.  Okay.  That is, this number, compared to that number.  Whether the number is stipulated or given in grams or moles or number of molecules,  it doesn't matter.  It can give you a ratio.  Okay.  Grams always have to be converted to moles first.  And then you get the mole ratio.  And the number of molecules that have a number of molecules is also mole ratio.  This number divided by that makes you 0.182.  Sorry, 181.  That is less than 0.2.  That means numerator, again, is the limited reagent.  You have all the numerators here.  So, there are numerators always the limited reagent.  You could have as well been the denominator.  In other words, it just happens to be the numerator.  Okay.  So, the numerator is the limited reagent.  In this case, this is a di-I-D pentoxide.  That means I take the number of limited reagent, okay,  and I convert that into the number of excess reagent.  Reactive.  How?  Multiply that in a ratio.  So, I take the number of the three units of this, I take with it 10 units of nitrogen oxide.  This times that gives you the number of nitrogen oxide mole  to mole with half the reactant in the process, 980.  This is the amount that I had in the beginning.  So, I take that, subtract this from it.  I have exactly 100 units left.  Units is nothing but the mole.  Okay.  So, this is the flow of these kind of aeroexercises.  You can see that, and hopefully that will not be too hard for you.  Okay.  So, let's do a couple of minutes.  Let's say there's no life that we have.  We apply to an actual problem.  So, the problem of exploring cars.  So, we've seen it in some of the movies.  We've seen it in some more of the left and right.  The same question we've been trying to solve before the problem.  It seems like it is extremely easy to realize.  Take a car and explore it.  Let's see if this really is that easy.  Okay.  What happens in this process is there's a chemical reaction.  One is exploding, gas will leave.  What does gas leave?  Mostly, captain.  That's captain.  And it reacts with oxygen.  That's the arrow that should go to the right.  And it generates C2O2 and H2O2.  This is combustion reaction.  And it's boundless being all red.  So, when does this explosion happen?  It happens when this, the meeting of O2, but half day, is very efficient.  So, if these chemicals are present in the right proportion,  they can very actively react and explore.  If there's not enough oxygen or not enough captain,  then the chemicals have to wait till they leave  and it slows down the process of the laserous process.  So, what I really want is I want this reaction to be complete.  I want the reagents to be present in the right proportions  and, in the optimum, react with conditions.  That's a long way to say that what I need is,  for each one mole of heptane, I need 11 moles of oxygen.  That's what I want in my tank.  So, let's look at that.  Here's a tank.  A typical tank is about 14 gallons, converted immediately to metric units.  It's about 53 liters.  This is a very rough definition.  Let's make it a 9-number, 53 liters is the approximate volume of a typical gas tank.  The density of heptane will need that,  this number right here, given in grams per mil.  And the density of oxygen is also given,  remember oxygen is a gas, okay, so that is 22.4 liters per mole.  And here's the reaction once again.  So, what do I need?  Well, I just said it.  What you want to do is you want to have quantities of oxygen and heptane  that are correct for reacting explosively.  That means 11 moles of oxygen for each one mole of heptane.  So, the ratio needs to be 11.  Okay, now let's see intuitively,  we think this is what your first gas would be.  I need as much heptane as possible to invest,  explosion time, completely fill the tank.  So, let's do that.  Let's assume the tank is a 24,  because I think in that case, more heptane means bigger explosion.  So, 53 liters, that's the whole thing, tank,  it's all full of heptane now.  How many moles of heptane is that?  That's a couple, I think.  So, the gas times the density converts a volume into a mass.  So, I multiply this by this number here.  This is the density, 0.684 grams per milliliter or kilograms per liter.  Remember, grams per milliliter is the same unit as kilograms per liters.  It has the same value of the density.  The value of the density is the same,  it's expressed in kilograms per liter, more grams per milliliter.  Or you can use that, and that's helpful because this is liter times liter,  so this crosses out, now you have the mass of heptane in the tank.  Okay, if the mass is expressed in kilograms,  I can express mass in moles by the value of the volume, mole mass.  So, watch out if you can do that correctly.  Usually, the mole mass is given in grams per mole,  but now I'm going to do kilograms per mole.  So, that means you divide it by a thousand.  The kilogram across out, what you get is moles of heptane.  Same calculation that we did.  You have the mass, you divide it by the mole mass.  The mass is known as kilograms divided by the mole mass,  and you can express in kilograms per mole.  Heptane is about 100 grams per mole, that means 0.1 kilograms per mole.  I find 3.6 times 10 to the second mole of heptane.  So, this is the amount of heptane that I got.  Okay, so I need relative to heptane,  11 times more oxygen for this we have to have.  Well, the problem is, I don't have any oxygen  because the thing is full.  There is no oxygen to react with.  The thing is completely full,  there is really no oxygen that can officially get to the heptane.  It's not in the same place.  This is a Q student pipe, that's R-I.  There's no oxygen there for all the molecules of heptane to react with.  So, this is absolutely not the situation that we're dealing with.  If you're dancing to school, you're very sick.  You're a lot of swell.  You can fire for yourself, you can make a car of fire,  you can make a car of speech, you can see, boom, swell.  Make sure you jump out before the falls.  So, what we need, what we need is we need more oxygen.  Okay, we need more oxygen to live here  because the reaction needs oxygen,  there's no oxygen in the present yet,  so I need to get some heptane out and put some oxygen in.  Okay, so let's assume for the following,  let's say I have only one liter of heptane there for me,  and let's assume for all kinds of purposes that the air is 100% hydrogen,  which is not sure.  Well, that's just a super-design.  One liter of heptane in tank,  one liter, it'll be the same calculation,  let's see how many moles are there.  One liter times the density gives me mass,  mass divided by mole, mass gives me moles.  I complete this calculation,  I find 6.8 moles of heptane,  not that the quantity of heptane.  And now I need to see how many moles are hydrogen.  So, let's see, if this is one mole of heptane,  the total of 53, so that means 52 liters is hydrogen,  and I assume now, again, all the air is hydrogen.  It's not sure what it's going to say.  So, let's just do that.  So, what I do now, I convert liters directly to mole,  because I have this conversion here, 22.4 liters per mole.  So, if I multiply 52 liters by this conversion factor,  I put liters at the bottom and mole on top,  I directly convert the mole.  I find 2.3 moles of hydrogen.  So, 52 liters of hydrogen gas, apparently, is 2.3 moles.  Well, due to the division in 2.3 moles by 6.8 moles of heptane,  it's 0.33, I need 11.  This is W plus.  So, even in a situation where 100% oxygen in the air,  one liter of heptane is maintained,  and it won't explode.  Very very unlikely.  So, when it actually does it, I want this thing to explode.  I'm not giving up.  So, I'm going to take almost all the heptane out of the tank.  Let's just say that I, you know,  I think it's almost completely filled with oxygen.  That means 53 liters is now oxygen,  and I'm going to convert how many moles that is  to the same calculation.  53, this conversion factor from up here,  this one at the bottom, this on top,  gives me 2.4 moles of oxygen.  So, I have 2.4 moles of oxygen in my tank.  How much heptane do I need to make this fully react?  Again, under the assumption, there's 100% oxygen in the air.  Okay, well, the ratio needs to be 1 to 11.  1 to 11.  What we need is 2.4 moles of oxygen.  I take the ratio of oxygen at the bottom,  11 moles of oxygen for each one mole of heptane.  So, effectively, you divide this thing by 11.  Okay?  This here is the number of moles of heptane.  Okay?  I multiply that by the molar mass.  That means how many grams I have.  So, this is the mass of heptane.  Okay?  And then I'm going to multiply that by,  let's see, by this density conversion.  So, I convert the mass into volume.  So, I divide it by the mass.  So, I'm going to do a free conversion here.  First, from number of oxygen to number of heptane that I need,  moles of heptane converted into grams of heptane,  and then mass, grams, converted into volume.  Through the density, I divide that by this.  You see, grams at the bottom across out,  I need a time to answer in milliliters,  32 milliliters of heptane.  32 milliliters is about a quarter of a total.  Okay?  1-eighth on a cup.  1-eighth on a cup is what you need approximately  to have 2.4 moles of oxygen to react explosively with heptane.  Okay?  So, that is basically what it says.  You need a small fill of gasoline on your tank  to make that thing explode.  And then you also need about 300 percent oxygen in here.  Then you can have a very, very strong flow of oxygen.  So, less is better, less is better.  Alright, so this is an example of calculations.  It uses mole ratios.  It also uses things like this,  volumes, densities, right?  These are things we have seen in the first part of this class.  We're getting close to the first couple of lectures.  We're all about densities, volumes, so forth.  This is a real-life situation.  You could do these calculations with the tools that we've talked about.  One more question.  Yes?  Yes, hold on a second.  There's a nice question here.  Okay?  So, does that necessarily mean that with, you know,  the degree of the particles on my tank,  it's really more density?  The question is, why does it explode in a movie?  Well, first of all, in a movie,  it's for the good sticks of dynamite,  to make it explode, right?  But particles can explode.  And typically, it comes to explode when you have a leak,  and stuff drips from the floor.  It's a small film with what you did,  and that's what it's set up.  It typically in itself,  it won't have conditions or something.  Okay, whoa!
Preparation_for_General_Chemistry_1P__Lecture_15__Fun__0_563.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__564_827.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__828_935.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__936_1079.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__1080_1631.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__1632_1967.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__1968_2807.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__2808_9443.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__9444_11135.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__11136_11219.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__11220_11279.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__11280_11363.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__11364_12155.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__12156_12335.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__12336_14543.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__14544_14771.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__14772_15719.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__15720_15839.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__15840_16055.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__16056_17183.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__17184_18875.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__18876_19943.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__19944_20159.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__20160_20927.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__20928_22403.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__22404_23291.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__23292_23351.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__23352_23723.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__23724_23855.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__23856_24995.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__24996_25127.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__25128_25175.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__25176_25475.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__25476_26291.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__26292_28595.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__28596_28643.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__28644_28703.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__28704_29555.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__29556_30587.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__30588_31379.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__31380_33239.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__33240_34043.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__34044_35255.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__35256_36035.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__36036_37895.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__37896_39575.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__39576_41339.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__41340_42167.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__42168_43031.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__43032_44039.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__44040_44231.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__44232_44963.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__44964_48731.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__48732_50831.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__50832_50879.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__50880_50999.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__51000_52175.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__52176_54131.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__54132_56099.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__56100_56231.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__56232_57107.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__57108_58739.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__58740_59795.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__59796_60011.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__60012_60983.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__60984_61859.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__61860_63251.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__63252_63395.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__63396_64235.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__64236_65639.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__65640_65987.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__65988_67559.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__67560_67619.png Preparation_for_General_Chemistry_1P__Lecture_15__Fun__67620_68057.png
Okay, finished. I'm going to see valves. I'm going to do a very important chemistry. And today I will not introduce anything new. I'm probably already sure. Oh, I'm going to bring up this guy. So, very cute, very cute. I love this. Lots of parents, if they have a cool baby, they give this baby an iPad. And then they're so happy that the baby after about figures out and press something out. Technically, a little child out is an amazing thing. The fact of the matter is, who wants more babies to do? It's not really the problem. As long as you train a little puppy something eventually, it will do that. You can train a dog to pee in his place, where we can train a baby to press this button. So, in any case, these are things that are within the capabilities of us being humans. These are things that we humans can do. Now, we humans can also do things like this. Who likes to go to the causes? I like what the great amount of people likes to go to the causes. I like it too. Every night, I do want to try to go to the causes. That's very cool. I can tell you that the go to the causes are much more complex than what the calculations we do here in this class. There's no comparison. This is orders of magnitude more complex, higher order of thinking, and the kind of exercises we do in this class. Mathematically, it's just very easy to show. Even this is more difficult. Yes it is. Yes it is. So, what I'm trying to do today is try to do a couple of exercises that are within the material that we're working with, and we'll try to break it down, and we'll try to put the pieces where they belong. The mathematical steps are a multiplication or division. So we have to be able to understand the problem and just do those operations correctly. So let me first start with balancing equations. Balancing equations is like a little puzzle. It is like a little puzzle. So let me give you an example. Here's one. What do you have to do? Well, I just have to make sure that on both sides there's the same amount of atoms of each kind. I see two here. There's only one there. I see three there. And there's two there, so evidently this is not balanced. Now what happens? This is two, and this is three. So there's no way you can make this equal by just multiplying this with, let's say, two and this with two. If I multiply this with two, I generate a total of six. Six oxygens. That means I need six oxygens here too. That means if I multiply three there, I get six. And because I multiply by two, I get four of these guys, and the four of those guys. This is it. So I'm just going back and forth until everything is the same inside. Another example. Strontium phosphorus. This four here, this one there. I need three here and two there, so this evidently is not correct. I can start trying to make one of the elements correct and then balance with the other one. Now what I'm trying to do now first, let me just show you why I do this in this order. I see a E four here, and I see a one here. This one means that I can put any number in front to match the missing correlation. So this is not the one I want to balance in the beginning. I want to make sure that the phosphorus is balanced first because I see, ah, there's no coefficient there. I can put any number in front to make a match. So I start with phosphorus in this case. I have four here. I need four on the other side, so I need to multiply this by two. Two times two is four. So I generate six of these guys, the six of them there. See why I'm starting with phosphorus, right? Now let's not do that. Okay, another type of question is this. This is just kind of like a huge chemical equation not to be so scared of them. I have two Y, and this Y is an unspecified compound. It reacts with Br2, and I have two times sodium bromide. What is the missing Y? If you just have to look at this, this question, unfortunately, if you look at the equation, you just put the pieces together, really. What I see here, I have sodium and bromine. Bromine is there, sodium is not there. So I need something with sodium, and there's an iodine there too. So I need sodium iodide. Sodium iodide. If I put this one here, let's check if this is correct. Two times sodium, excuse me, two sodiums, two times iodine gives me I2. So this is a balance equation. You know, this is really actually no puzzle. This is a puzzle, I think, that falls within our union chain of deltas. There's nobody that is so challenged that they cannot solve this puzzle. The only thing is to overcome your resistance for a chemical equation. That's it. As soon as you take out the stack, then you can do the solving. Let's do this one here. Why? If you set the fill out, what is why? I see here hydrogen sulfur. Sulfur is here, hydrogen is there. And it's also oxygen is there. So oxygen is missing. Okay? So this is most likely an oxygen. So let's say this is O2. Let's see if I put O2 in there, if I actually balance the equation. Three times O2 is six octagons. Two times two is four. Two times one is two. Two plus four is six. So that means six octagons. That's right. The other one is four hydrogens, four hydrogens, and two solvers, two solvers. So this is balance, this is balance. Okay, so make sure you're just comfortable with these equations. They're little puzzles and they can be solved. All right, now let's do this. So this is a little bit of a visual illustration of the actual problem. Sometimes in the building house, if you take a question from the abstract, which is a text, into something that is actually completely an object, see the molecules and then it is built into an act. And this is the reaction that we're looking at. This should be an arrow. This is H2, and O2 is these guys. So the red ones are the oxygens, the gray ones are hydrogens, and the fourth one is water. Each water has one oxygen, two hydrogens. So the one to make a water molecule here, I need at least two of these hydrogens and one oxygen. I see one, two, three, four, five, six oxygens, so I could tentatively make six water molecules. I count six oxygens, I can make six water molecules. Okay, and that means if I can make six, I need for each water molecule two hydrogens. So each hydrogen molecule contributes two hydrogens. So if I need six water molecules, I need six of these pairs. One, two, three, four, five, six. So if I use those six plus those guys, I have these two left. So what is the situation after the reaction happens? Choose from this. I just determined that if this react reacts, I can form six water molecules out of three O2s and six H2s, and then I have two H2s left, two hydrogen molecules left. They will not react because they're running out of red spheres. So this one is not correct because I have an accessory agent which is H2. This one is also not correct because I said I could make six. This one makes three, also not correct. Look, this guy here can still react. This oxygen can react with this guy, the other oxygen with this guy can form water molecules. One, two, three, four, five, six plus two hydrogens, this must be the right one. Okay, so again, it is basically a puzzle. You just lose the idea to click, and you just have to determine how many do you form, and what do you have left and what do you have left and what. That's the problem. Okay. Now look at this. This is an interesting one. This is a sample of a compound that contains chlorine and oxygen. So this is a compound that has two elements in it, C, L, and R. Nothing else. It will react with an excess of H2. So this is the accessory agent, the other element, the other compound that's unknown at this point is the limited reagent, and they will produce two products, H, C, L, and H2O. The quantities are given in grams. So determine the empirical formula of the compound. Now, if you read this the first time, you may say, what's going on? How can I determine the empirical formula for this piece of information? That's the first response. The second response is, okay, let's first look at that. Let's see what that is. You want to visualize this problem, okay? The way I visualize it, I'm going to write a chemical equation. I'm not sure what happens with this error. Sorry, it should just be an error to the right. Apparently there's a compound that contains C, L, and O, but I don't know the substance because that's unknown. That's what I have to determine, right? So I call it X and Y. I know it reacts with H2. I also know there's a lot of H2, so H2 is not a limiting reagent. If this is a limiting reagent, okay? It produces H, C, L, and H2O. Okay, so that's that number. And I have to determine apparently what this X and Y is. How do I get that? Well, let's have a look at this problem a little closer. I see that each mole of C, L, X, or Y can produce how many HCl's? If there's X times C, L in there, if I have one mole of this, I can make X mole of HCl, right? Because each chlorine gives rise to one HCl mole. So if I have X4H in my compound, I can make XHCl's. Same for oxygen. Each water molecule has one oxygen. The oxygen must come from this. There's no other source. So if I have Y oxygen atoms, I can make one-eight quarters. What about the H? Well, I just said, no one but eight. There's enough H to take care of this. The limiting reaction is this. So the amount of HCl and H2O is determined by this compound. That means I can make X mole of HCl, Y mole of H2O. All right? Now, that is almost already the problem, because I owe us a solution to the problem, because I have in principle the quantities and moles of these guys. It's given in grams, but I know right now that I can convert grams in a flash into moles. Okay? By the time I'm done with moles. So that's what I'm going to do. I'm going to see how many moles of HCl did I generate? How many moles of H2O did I generate? The ratio between the two should be the ratio between X and Y. Okay. That's what basically what this line says. If I determine how many moles of HCl I generate, how many moles of H2O I generate, then I know the ratio between X and Y. And the ratio between X and Y is the interval. Okay? Okay, let's do this. So what I have to do is I have to calculate how many moles of this I have and how many moles of that. That is something we can do. Okay? Grams converting to moles by dividing by the mole of X. That's your conversion pattern. You don't know yet that you have to divide. You always do this conversion and check the units this has to cross out. This conversion will give you the number of moles of HCl generated. I find 6.3 or 6.4 times the number of moles of this compound. Let's assume everything happens in water. And this thing here is hydrochloric acid. How about the other guy? Okay, so this HCl corresponds to this many moles of chlorine atoms, right? Because each molecule has only one chlorine atom. So this is the amount of chlorine. All right? This is the quantity of H2O. I quickly convert that to moles by dividing by the mole of H2O. This is a division. What I get is 2.24 times 10 to the minus 2 moles of H2O. Each H2O molecule contributes one oxygen. That means I have 2.24 times 10 to the minus 2 moles of oxygen. So this is then X moles of this, Y moles of that. So this is X is Y. I have to calculate the ratio between the two. So that means I can find the empirical ratio between chlorine and oxygen. Chlorine I have the quantity of, oxygen I have the quantity of. The ratio between the two is the ratio between the subscripts. How many moles of C I have relative to the amount of moles of oxygen? So if I divide one by the other, let's just put oxygen on top. It doesn't have to be a quantity, right? Just choose one. Oxygen on top, divide that by the amount of moles of chlorine. The ratio is 3.5. That means for each chlorine I have 3.5 oxygens. Right? For each chlorine I have 3.5 oxygens. That's what this calculation says. We also know that in the empirical formula, we don't have fractions. It should be an integer. So it's not close to an integer number. Is that a problem? Not necessarily. I could just multiply it with a number so that I can make an integer. So let's look at that. I have here for each one Cl, 3.5 oxygens. That's what I just determined. For each one Cl, for each one mole of Cl, I have 3.5 moles of oxygen. If I multiply this whole unit by two, I get this. Cl2 of 7. That is the empirical formula. It's not the molecular formula yet. That's not a question. The question is determining the empirical formula. And the answer is right here. So this is a slightly different question. But again, with the same kind of thinking. How many atoms do I have? How many moles do I have? And how do they relate to one another? If I have this much of this, how much do I have to be audible? That is the kind of thinking I have for applying the multiply. I have a question. Why do you multiply it by two? The question is why do you multiply it by two? Because an empirical formula should give you the substance in integers. This is not an integer. This is 3.5. You have to do that. You have to multiply it with the lowest multiplier that gives you the degree. And that is what we do here. The lowest multiplier is two because two times 3.5 is seven. That is an integer. And in the process, I have two here as well. The empirical formula CO2, 07. Yes, another question. I have a question. I said hydrogen is an excess. I did not have to worry about the high. But for each molecule of HCl, I need one atom of H. One hydrogen. For each molecule of oxygen of water, I need two atoms of hydrogen. So it depends on if you look at this guy or this guy. So there are different ratios. You need more H than you need oxygen and CO. But fortunately, there is an abundance of Hs in the process. So I do not have to worry about it. The pressure is not about that. The pressure is typically about that. And then we just look at how many Hs in that. It needs to be tonic energy. And how much we need. And how much H we have consumed. You can do that calculation as well. Okay. So let's do the following. Another reaction. Remember, these typos are arrows. C6 H14 is converted into C2 H4. That is a reaction. There is also other stuff that I do not have to worry about. Basically, this means that I have one mole of this that is converted into one mole of that. The ratio is one to one. If the yield of this process is 42.5%, what mass of this starting material, which is called hexane, must be there to produce 481 grams of the product, which is called ethylene. So, in other words, your boss tells you, I need from you 481 grams of ethylene. Okay? And he also tells you what. This reaction is on very efficient. Efficiency is exactly 42.1, 42.5. And I want you to determine how much hexane do I need to pour. Can you give me that answer finally? He said, ah, I can do that. Come on, I need only three. All right. So what you do is this. Okay? You would like to know how many moles of the product do I need? How many moles of the product do I need? Again, grams, you want to move very quickly, convert grams all the way to moles. So, something like a countable number, which is moles. 48, 481 grams divided by the molar mass of ethylene. Okay? Which is 28.05. Says 70.1 moles of ethylene. The boss apparently needs 70.1 moles of ethylene. That's what he needs. So that means if you have 70.1 moles of ethylene, how much starting material do we need? Well, the reaction says it reacts one to one. One mole of hexane, the starting material, the reagent produces one mole of this. Okay? So, if this reaction were to be one hundred percent efficient, you would need 70.1 moles of the starting material to react one to one. 70.1 moles in, 70.1 moles out. Okay? However, the efficiency is 42.5. So, you have to take that into consideration. Apparently, you need more than 70.1 because efficiency is rather large. So, how do you do that? Okay? So, this is what I just said. So, take into account the yield. The yield expressed in a fraction times the amount that you need will be 70.1 moles. Okay? You need 70.1 moles. That should be the outcome. Okay? You need 70.1 moles. But you need to put more in because you multiply something as small as one. Okay? So, the fraction corresponding to this percentage is 0.425 times the amount that you have to put in should give you the amount that you have to give to your supervisor. So, that means the amount that I need, I can quickly calculate, 17.1 moles divided by this fraction 40.2 moles is the amount you have to start with. Now, this is the amount of moles. The question also says, I need it in grams. So, what you have to do is, you have to multiply this by the molar mass to get the amount of grams in grams. Is this set clear? Does that make sense? Alright. So, the multiplication of the molar mass you can do for yourself when you have the answer. So, we just move on to this. This question. This question is the same kind of question, but the phrasing is a little different. Now, don't be shaken by this. It is again the same kind of thinking. And the reason why I like this example is it forces you to really think about these things as countable objects. Some of this, some of that, the reactometry and so forth. Okay, so here's another reaction. Methane reacts with oxygen, combustion reaction with methane. CO2 and 2H2O comes out. I have to determine the number of excess reagent. So, excess reagent means what is left in the pot after this reaction is over. Okay? How many do I have in the beginning? Well, it says I have 70 molecules of methane and 142 of oxygen. The funny thing about this question is it doesn't say how many moles you have or how many grams you have. Okay? That is another problem. That's the same thing. Another molecule suggests another molecule. But it is specified in an actual number for a mole that doesn't matter. Irrigate with the numbers. Okay, so what I have to do is to determine which one of these reagents is the limited area. In order to do that, I play the same trick as before. I want to know what is the mole ratio between these two guys according to the equation, the balanced equation. It says you need two moles of oxygen for each one mole of methane. And that means the ratio of 0.5. If you put methane on top, the ratio will be 0.5. The next thing I'm going to do, so this is the mole ratio. Mole ratio. I want to determine the mole ratio that I actually have. Again, the number of moles is not given. The number of molecules is given. But what is a mole? A mole is a number. It is 6.05. So, we're at 6.02, the time is 6.33. So, I think you can convert this to moles if you want. You multiply this by moles, by a mole number, this by a mole number. But the ratio will change. Because if you multiply this by a mole, this by a mole, you divide them out, you just see what I'm saying. The ratio is determined by the proportionality between the two. Whether you have a mole of it, or just a countable number of molecules on your fingers, the ratio is still the same. So, I can determine the mole ratio simply by dividing the quantities of these numbers. Of the number of molecules that I have. Okay? So, let's just move on with this. So, 70 molecules of CH4, 142 molecules of O2. So, this is the actual mole ratio. Not given in moles, but in just molecules. Because the ratio between molecules and the ratio between moles is the same. Okay? I can convert this to moles if you want. It doesn't matter, because you convert also this one to more. The ratio between the two is 0.49. 0.49 is a little bit smaller than 0.5. I think the numerator is the limiting reagent. I eat the molecule of one dot first. You will burn through the 70 sooner than you burn through the 140. Okay? So, it's a smaller than 0.5. That means CH4, methane, is the limiting reagent. That means the other guy is the excess reagent. Okay? This O2. That means I have something left in the pot after the reaction has completed. Okay? So, how many molecules of oxygen did react? Same kind of calculation. 70 molecules of methane were there. That's the limiting reagent. I will convert the number of reactive methane to the number of reactive oxygen through the mole ratio. Okay? Two molecules of oxygen for each one molecule of CH4. That's what the equation says. That's the ratio between the two. For each one molecule of methane, any two molecules of oxygen. So, each one molecule of methane, two molecules of oxygen. This one goes at the bottom because I want to cross this out. I want to go to the number of molecules of oxygen that has reacted. 70 times 2 equals 140 molecules of oxygen have reacted in this process. Okay? So, how many do I have left? Yes, you have 142 minus 140, two molecules of oxygen. Now, mind you, this is exactly what we did before. Previously, we did it at quantities expressed in moles. Now, we do it in the physical number of molecules. But it's exactly the same because you work with ratios. You work with ratios. So, you're going to replicate them exactly the same. I'm going to do another one which is virtually the same. Just to get used to this. The same kind of question, a different reaction. This is also open-bush reaction. It's a different kind of molecule. But the derivative of this should be the same. Okay? Calculate the number of excess reagent molecules after this reaction has proceeded. I'll do the same thing again. First, the mole ratio according to the equation. The equation says two moles of this reagent for each 15 moles of oxygen that ratio is 21.36. That's what I call the Bellos ratio. The ratio stipulated by the Bellos chemical equation. Then, I want to see what I actually have. What is the ratio of the chemicals to reagents that I actually have? Well, here are the numbers that I actually have. Okay? The same kind of calculation. 74 divided by 585 equals 0.126. This number? Less than that. That means the numerator is going to be my limiting reagent. Okay? So, C5H10 is the limiting reagent. You can always determine how many molecules have reacted. All the limiting reagent react, and when the limiting reagent reacts, it takes with it, in this ratio, the oxygen. That means it's compared by how many oxygen's have reacted in the process. So, that conversion is, again, number of molecules of limiting reagent is reacting. It takes with it, for each two of this, 15 molecules of oxygen. That's the mole ratio of the two molecules. Okay? So, 74 molecules of this reagent react in this ratio. Each two molecules reacting take with it 15 molecules of oxygen. That means 55 oxygen molecules have reacted. The conversion here is from number of molecules of C5H10 to number of molecules of O2. O2 goes on top. This guy's going at the bottom. This is what I'm always doing. We need to convert from one quality into the next. How many did I have in the beginning? 585. This many have reacted. So, 555 minus this 30 molecules of oxygen are left. They cannot react. Why? Because there's no other reagent anymore. It can't go on. Okay. I think we've got to run for it. I do one more of this, and then we'll do a final thing accepting SSI. Okay. So, we do it all quick. Another reagent is safe. Unlike repetition, it really gets you into the mood. And these, like I said, dissolve white foam. Okay. So, another compound reacting with an oxygen oxide producing these two compounds. And again, I've got the excess reagent. So, he is the truncable limited reagent. What you do is look at the mole ratio before we get up to the chemical equation. Two to ten, at least, points. You compare that with the actual number that you have. Okay. That is, this number, compared to that number. Whether the number is stipulated or given in grams or moles or number of molecules, it doesn't matter. It can give you a ratio. Okay. Grams always have to be converted to moles first. And then you get the mole ratio. And the number of molecules that have a number of molecules is also mole ratio. This number divided by that makes you 0.182. Sorry, 181. That is less than 0.2. That means numerator, again, is the limited reagent. You have all the numerators here. So, there are numerators always the limited reagent. You could have as well been the denominator. In other words, it just happens to be the numerator. Okay. So, the numerator is the limited reagent. In this case, this is a di-I-D pentoxide. That means I take the number of limited reagent, okay, and I convert that into the number of excess reagent. Reactive. How? Multiply that in a ratio. So, I take the number of the three units of this, I take with it 10 units of nitrogen oxide. This times that gives you the number of nitrogen oxide mole to mole with half the reactant in the process, 980. This is the amount that I had in the beginning. So, I take that, subtract this from it. I have exactly 100 units left. Units is nothing but the mole. Okay. So, this is the flow of these kind of aeroexercises. You can see that, and hopefully that will not be too hard for you. Okay. So, let's do a couple of minutes. Let's say there's no life that we have. We apply to an actual problem. So, the problem of exploring cars. So, we've seen it in some of the movies. We've seen it in some more of the left and right. The same question we've been trying to solve before the problem. It seems like it is extremely easy to realize. Take a car and explore it. Let's see if this really is that easy. Okay. What happens in this process is there's a chemical reaction. One is exploding, gas will leave. What does gas leave? Mostly, captain. That's captain. And it reacts with oxygen. That's the arrow that should go to the right. And it generates C2O2 and H2O2. This is combustion reaction. And it's boundless being all red. So, when does this explosion happen? It happens when this, the meeting of O2, but half day, is very efficient. So, if these chemicals are present in the right proportion, they can very actively react and explore. If there's not enough oxygen or not enough captain, then the chemicals have to wait till they leave and it slows down the process of the laserous process. So, what I really want is I want this reaction to be complete. I want the reagents to be present in the right proportions and, in the optimum, react with conditions. That's a long way to say that what I need is, for each one mole of heptane, I need 11 moles of oxygen. That's what I want in my tank. So, let's look at that. Here's a tank. A typical tank is about 14 gallons, converted immediately to metric units. It's about 53 liters. This is a very rough definition. Let's make it a 9-number, 53 liters is the approximate volume of a typical gas tank. The density of heptane will need that, this number right here, given in grams per mil. And the density of oxygen is also given, remember oxygen is a gas, okay, so that is 22.4 liters per mole. And here's the reaction once again. So, what do I need? Well, I just said it. What you want to do is you want to have quantities of oxygen and heptane that are correct for reacting explosively. That means 11 moles of oxygen for each one mole of heptane. So, the ratio needs to be 11. Okay, now let's see intuitively, we think this is what your first gas would be. I need as much heptane as possible to invest, explosion time, completely fill the tank. So, let's do that. Let's assume the tank is a 24, because I think in that case, more heptane means bigger explosion. So, 53 liters, that's the whole thing, tank, it's all full of heptane now. How many moles of heptane is that? That's a couple, I think. So, the gas times the density converts a volume into a mass. So, I multiply this by this number here. This is the density, 0.684 grams per milliliter or kilograms per liter. Remember, grams per milliliter is the same unit as kilograms per liters. It has the same value of the density. The value of the density is the same, it's expressed in kilograms per liter, more grams per milliliter. Or you can use that, and that's helpful because this is liter times liter, so this crosses out, now you have the mass of heptane in the tank. Okay, if the mass is expressed in kilograms, I can express mass in moles by the value of the volume, mole mass. So, watch out if you can do that correctly. Usually, the mole mass is given in grams per mole, but now I'm going to do kilograms per mole. So, that means you divide it by a thousand. The kilogram across out, what you get is moles of heptane. Same calculation that we did. You have the mass, you divide it by the mole mass. The mass is known as kilograms divided by the mole mass, and you can express in kilograms per mole. Heptane is about 100 grams per mole, that means 0.1 kilograms per mole. I find 3.6 times 10 to the second mole of heptane. So, this is the amount of heptane that I got. Okay, so I need relative to heptane, 11 times more oxygen for this we have to have. Well, the problem is, I don't have any oxygen because the thing is full. There is no oxygen to react with. The thing is completely full, there is really no oxygen that can officially get to the heptane. It's not in the same place. This is a Q student pipe, that's R-I. There's no oxygen there for all the molecules of heptane to react with. So, this is absolutely not the situation that we're dealing with. If you're dancing to school, you're very sick. You're a lot of swell. You can fire for yourself, you can make a car of fire, you can make a car of speech, you can see, boom, swell. Make sure you jump out before the falls. So, what we need, what we need is we need more oxygen. Okay, we need more oxygen to live here because the reaction needs oxygen, there's no oxygen in the present yet, so I need to get some heptane out and put some oxygen in. Okay, so let's assume for the following, let's say I have only one liter of heptane there for me, and let's assume for all kinds of purposes that the air is 100% hydrogen, which is not sure. Well, that's just a super-design. One liter of heptane in tank, one liter, it'll be the same calculation, let's see how many moles are there. One liter times the density gives me mass, mass divided by mole, mass gives me moles. I complete this calculation, I find 6.8 moles of heptane, not that the quantity of heptane. And now I need to see how many moles are hydrogen. So, let's see, if this is one mole of heptane, the total of 53, so that means 52 liters is hydrogen, and I assume now, again, all the air is hydrogen. It's not sure what it's going to say. So, let's just do that. So, what I do now, I convert liters directly to mole, because I have this conversion here, 22.4 liters per mole. So, if I multiply 52 liters by this conversion factor, I put liters at the bottom and mole on top, I directly convert the mole. I find 2.3 moles of hydrogen. So, 52 liters of hydrogen gas, apparently, is 2.3 moles. Well, due to the division in 2.3 moles by 6.8 moles of heptane, it's 0.33, I need 11. This is W plus. So, even in a situation where 100% oxygen in the air, one liter of heptane is maintained, and it won't explode. Very very unlikely. So, when it actually does it, I want this thing to explode. I'm not giving up. So, I'm going to take almost all the heptane out of the tank. Let's just say that I, you know, I think it's almost completely filled with oxygen. That means 53 liters is now oxygen, and I'm going to convert how many moles that is to the same calculation. 53, this conversion factor from up here, this one at the bottom, this on top, gives me 2.4 moles of oxygen. So, I have 2.4 moles of oxygen in my tank. How much heptane do I need to make this fully react? Again, under the assumption, there's 100% oxygen in the air. Okay, well, the ratio needs to be 1 to 11. 1 to 11. What we need is 2.4 moles of oxygen. I take the ratio of oxygen at the bottom, 11 moles of oxygen for each one mole of heptane. So, effectively, you divide this thing by 11. Okay? This here is the number of moles of heptane. Okay? I multiply that by the molar mass. That means how many grams I have. So, this is the mass of heptane. Okay? And then I'm going to multiply that by, let's see, by this density conversion. So, I convert the mass into volume. So, I divide it by the mass. So, I'm going to do a free conversion here. First, from number of oxygen to number of heptane that I need, moles of heptane converted into grams of heptane, and then mass, grams, converted into volume. Through the density, I divide that by this. You see, grams at the bottom across out, I need a time to answer in milliliters, 32 milliliters of heptane. 32 milliliters is about a quarter of a total. Okay? 1-eighth on a cup. 1-eighth on a cup is what you need approximately to have 2.4 moles of oxygen to react explosively with heptane. Okay? So, that is basically what it says. You need a small fill of gasoline on your tank to make that thing explode. And then you also need about 300 percent oxygen in here. Then you can have a very, very strong flow of oxygen. So, less is better, less is better. Alright, so this is an example of calculations. It uses mole ratios. It also uses things like this, volumes, densities, right? These are things we have seen in the first part of this class. We're getting close to the first couple of lectures. We're all about densities, volumes, so forth. This is a real-life situation. You could do these calculations with the tools that we've talked about. One more question. Yes? Yes, hold on a second. There's a nice question here. Okay? So, does that necessarily mean that with, you know, the degree of the particles on my tank, it's really more density? The question is, why does it explode in a movie? Well, first of all, in a movie, it's for the good sticks of dynamite, to make it explode, right? But particles can explode. And typically, it comes to explode when you have a leak, and stuff drips from the floor. It's a small film with what you did, and that's what it's set up. It typically in itself, it won't have conditions or something. Okay, whoa!
formula. It's not the molecular formula yet. pressure is not about that. The pressure is how many Hs in that. It needs to be tonic energy. that doesn't matter. Irrigate with the numbers. is a mole? A mole is a number. It is 6.05. So, because you work with ratios. You work with So, I take the number of the three units of this, is the approximate volume of a typical gas tank. the density. The value of the density is the as kilograms divided by the mole mass, and you