Ethyl alcohol $(C_{2} H_{5} O H)$ is prepared industrially by the reaction of ethylene $(C_{2} H_{4})$ with water. What is the per cent yield of the reaction if $9.6$ g of ethylene gives $4.7$ g of ethyl alcohol?(Round-off the answer to the nearest integer).
$C_{2} H_{4} (g) + H_{2} O (l) ⟶ C_{2} H_{5} O H (l)$

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Solution

The molar masses of ethylene and ethyl alcohol are $28$ g/mol and $46$ g/mol respectively.
$1$ mole ( $28$ g) of ethylene gives $1$ mole ( $46$ g) of ethanol.
Hence, the theoretical yield is $9.6 \times \frac{46}{28} = 15.77$ .
The percentage yield is $\frac{15.77}{46} \times 100 = 29.8 %$ .

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