# Chapter 3 - Chemical Equations and Reaction of Stoichiometry - Exercises - Limiting Reactant - Page 108: 34

(a) $S_8$ is the limiting reactant. (b) 67.5 g of $S_2Cl_2$ (c) 35.6 g of $Cl_2$

#### Work Step by Step

- Calculate or find the molar mass for $S_8$: $S_8$ : ( 32.07 $\times$ 8 )= 256.56 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$32.0 \space g \times \frac{1 \space mole}{ 256.56 \space g} = 0.125 \space mole$$ - Calculate or find the molar mass for $Cl_2$: $Cl_2$ : ( 35.45 $\times$ 2 )= 70.90 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$71.0 \space g \times \frac{1 \space mole}{ 70.90 \space g} = 1.00 \space mole$$ Find the amount of product if each reactant is completely consumed. $$0.125 \space mole \space S_8 \times \frac{ 4 \space moles \ S_2Cl_2 }{ 1 \space mole \space S_8 } = 0.500 \space mole \space S_2Cl_2$$ $$1.00 \space mole \space Cl_2 \times \frac{ 4 \space moles \ S_2Cl_2 }{ 4 \space moles \space Cl_2 } = 1.00 \space mole \space S_2Cl_2$$ Since the reaction of $S_8$ produces less $S_2Cl_2$ for these quantities, it is the limiting reactant. - Calculate or find the molar mass for $S_2Cl_2$: $S_2Cl_2$ : ( 35.45 $\times$ 2 )+ ( 32.07 $\times$ 2 )= 135.04 g/mol - Using the molar mass as a conversion factor, find the mass in g: $$0.500 \space mole \times \frac{ 135.04 \space g}{1 \space mole} = 67.5 \space g$$ - Find the amount of $Cl_2$ consumed. $$0.125 \space mole \space S_8 \times \frac{ 4 \space moles \ Cl_2 }{ 1 \space mole \space S_8 } = 0.500 \space mole \space Cl_2$$ $$0.500 \space mole \times \frac{ 70.90 \space g}{1 \space mole} = 35.4 \space g$$ Excess = Initial - Consumed = 71.0 g - 35.4 g = 35.6 g

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