Answer
$82.3\%$
Work Step by Step
$C_{2}H_{4}$ is the limiting reactant. The theoretical yield of ethanol is the mass of ethanol produced based on the amount of $C_{2}H_{4}$.
Theoretical yield=$10.0\times10^{3}\,g\times\frac{1\,mol\,C_{2}H_{4}}{28.05\,g}\times\frac{1\,mol\,CH_{3}CH_{2}OH}{1\,mol\,C_{2}H_{4}}\times\frac{46.07\,g}{1\,mol\,CH_{3}CH_{2}OH}=16.4\,kg$
$\text{% yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times100\%=\frac{13.5\,kg}{16.4\,kg}\times100\%=82.3\%$
