Question

$Hg$ and $I_2$ are allowed to react to form $H{g}_2{I}_2$(assume no other side products to be formed). 50 g of $Hg$ is exposed to excess $I_2$ and 65 g of $H{g}_2{I}_2$ is obtained. Calculate the percentage yield of the reaction.
(Molar mass of $Hg$ and $I$ are 200 g/mol and 127 g/mol respectively)
$2Hg+I_2\to H{g}_2{I}_2$

• A. 20.5 %
• B. 79.5 %
• C. 40.8 %
• D. 59.6 %

Answer 1

The correct option is B 79.5 %

$2Hg+I_2\to H{g}_2{I}_2$
Number of mole of $Hg$ = $\frac{\text{Given mass}}{\text{Molar mass}}=\frac{50}{200}=0.25\text{mol}$
1 mol of $H{g}_2{I}_2$ is produced from 2 mol of $Hg$
Since, $I_2$ is taken in excess
Theoretically, 0.25 mol of $Hg$ will produce 0.125 mol of $H{g}_2{I}_2$
Theoretical yield = number of moles of $H{g}_2{I}_2\times$ molar mass of $H{g}_2{I}_2$
Theoretical yield $=0.125\times 654=81.75$ g
Now, we know that
$\text{\% yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100=\frac{65}{81.75}=79.5\%$
Therefore, the percentage yield of the given reaction is 79.5 %