Question

$58\text{g}$ of butane on combustion gave $158.4\text{g}$ of $C{O}_2$. The Reaction is given below,
$2C_4{H}_{10}+13O_2\to 8C{O}_2+10H_{2}O$
Calculate the % yield of the combustion reaction.

• A. $50$
• B. $40$
• C. $90$
• D. $80$

Answer 1

The correct option is C $90$

$2C_4{H}_{10}+13O_2\to 8C{O}_2+10H_{2}O$
Number of moles of butane $=\frac{58}{58}=1\text{mol}$
As per stoichiometry,
2 mol of butane produces 8 mol of $C{O}_2$.
1 mol of butane will produce $\frac{8}{2}=4$ mol of $C{O}_2$.
So, theoretical yield $=4$ $\times$ molar mass of $C{O}_2$
theoretical yield $=4\times 44=176\text{g}$
$\text{\% yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$
$\text{\% yield}=\frac{158.4}{176}\times 100=90\%$