Answer
$61.7\%$
Work Step by Step
Ethanol is the limiting reactant. The theoretical yield of ethyl propionate is the mass of ethyl propionate produced based on the amount of $C_{2}H_{4}$.
Theoretical yield=$255\,g\times\frac{1\,mol\,CH_{3}CH_{2}OH}{46.07\,g}\times\frac{1\,mol\,ethyl\,propionate}{1\,mol\,CH_{3}CH_{2}OH}\times\frac{102.13\,g}{1\,mol\,ethyl\, propionate}=565.3\,g$
$\text{% yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times100\%=\frac{349\,g}{565.3\,kg}\times100\%=61.7\%$
