Question

$4.14$ g pure lead was dissolved in nitric acid and was made to react with $HCl,C{l}_2$ and $N{H}_{4}Cl$ to convert lead completely into $(N{H}_4{)}_{2}PbC{l}_6$. However, only $2.28$ g of $(N{H}_4{)}_{2}PbC{l}_6$ was actually produced. The percentage yield of the product is :

• A. $75$
• B. $35.5$
• C. $50$
• D. $25$

Answer 1

The correct option is D $25$

The atomic mass of $Pb$ is $207$ g/mol.

The number of moles in $4.14$ g of $Pb$ are $\frac{4.14}{207}=0.02$ moles.

The Molar mass of $(N{H}_4{)}_{2}PbC{l}_6$ is 456 g/mol.

$2.28$ g of $(N{H}_4{)}_{2}PbC{l}_6$ corresponds to $\frac{2.28}{456}=0.005$ moles.

The percentage yield $=\frac{0.005\times 100}{0.02}=25$ %