# Calculate the volume of hydrogen required for complete hydrogenation of 0.25 dm^3 of ethyne at STP?

Jul 18, 2015

You'd need ${\text{50 dm}}^{3}$ of hydrogen.

#### Explanation:

Like with any stoichiometry problem, the key tool you have at your disposal is the mole ratio.

However, an interesting thing takes place when you're dealing with gases that are under the same conditions for pressure and temperature. In such cases, the mole ratio becomes the volume ratio.

The idea is that, since you're dealing with two ideal gases, you can use the ideal gas law equation to write

$P \cdot {V}_{1} = {n}_{1} \cdot R T$ $\to$ for ethyne;

$P \cdot {V}_{2} = {n}_{2} \cdot R T$ $\to$ for hydrogen.

The pressure and the temperature are the same for both gases, since the reaction presumably takes place at STP.

If you divided these two equations, you'll get

$\frac{\cancel{P} \cdot {V}_{1}}{\cancel{P} \cdot {V}_{2}} = \frac{{n}_{1} \cdot \cancel{R T}}{{n}_{2} \cdot \cancel{R T}}$

This is equivalent to

${n}_{1} / {n}_{2} = {V}_{1} / {V}_{2}$

The mole ratio is equivalent to the volume ratio.

All you need now is the balanced chemical equation for the reaction, which looks like this

${C}_{2} {H}_{2 \left(g\right)} + \textcolor{red}{2} {H}_{2 \left(g\right)} \to {C}_{2} {H}_{6 \left(g\right)}$

Notice that you have a $1 : \textcolor{red}{2}$ mole ratio between ethyne and hydrogen gas. This means that the reaction needs twice as many moles of hydrogen gas than of ethyne.

This translates into volumes as well. In other words, the volume of hydrogen gas must be twice as big as the volume of ethyne.

$0.25 \cancel{\text{dm"^3"ethyne") * (color(red)(2)" dm"^3 "hydrogen")/(1cancel("dm"^3 "ethyne")) = color(green)("0.50 dm"""^3"hydrogen}}$