Question

Ethylene oxide $\left(C_2{H}_{4}O\right)$ is manufactured by burning ethylene $\left(C_2{H}_4\right)$ in air. If 60 g of ethylene oxide is obtained from 42 g of ethylene, calculate the % yield of ethylene oxide.

• A. 90.90 %
• B. 78.38 %
• C. 46.69 %
• D. 14.25 %

Answer 1

The correct option is A 90.90 %

Ethylene oxide is prepared from ethylene by the following balanced reaction:
$2C_2{H}_4+O_2\to 2C_2{H}_{4}O$
Moles of $C_2{H}_4=\frac{42}{28}=1.5$ moles
2 moles of $C_2{H}_4$ produces 2 moles of $C_2{H}_{4}O$
So, 1.5 moles produces = 1.5 moles of $C_2{H}_{4}O$
$\text{Mass of}{C}_2{H}_{4}O\text{obtained}=\text{moles}\times \text{molar mass of}{C}_2{H}_{4}O=1.5\times 44=66g$
Obtained (experimental) amount of $C_2{H}_{4}O=60$ g (given)
$\text{\% yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$
$=\frac{60}{66}\times 100=90.90\%$