Question

Ethylene oxide $\left(C_2{H}_{4}O\right)$ is manufactured by burning ethylene $\left(C_2{H}_4\right)$ in air. If 4.95 g of ethylene oxide is obtained from 4.2 g of ethylene, calculate the % yield of ethylene oxide.
$C_2{H}_4+O_2\to C_2{H}_{4}O$

• A. 90.90%
• B. 75.00%
• C. 46.69%
• D. 14.25%

Answer 1

The correct option is B 75.00%

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 $C_2{H}_4=\frac{4.2}{28}=0.15$ moles
2 moles of $C_2{H}_4$ produces 2 moles of $C_2{H}_{4}O$
So, 0.15 moles produces = 0.15 moles of $C_2{H}_{4}O$
$\text{Mass of}{C}_2{H}_{4}O\text{obtained}=\text{moles}\times \text{molar mass of}{C}_2{H}_{4}O=0.15\times 44=6.6g$
Obtained (experimental) amount of $C_2{H}_{4}O=4.95$ g (given)
$\text{\% yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$
$=\frac{4.95}{6.6}\times 100=75\%$