Question

The mass of $AgCl$ obtained by the reaction of $169.8g$ of $AgN{O}_3$ with $29.2g$ of $NaCl$ according to following equation is $27.2g$.
$AgN{O}_3+NaCl\to AgCl+NaN{O}_3\left(\text{not balanced}\right)$
Calculate the percentage yield.

$($Molar mass of $AgN{O}_3$ is $169.8g/mol$, molar mass of $NaCl$ is $58.4g/mol$ and molar mass of $AgCl$ is $143.3g/mol)$

• A. 33.33%
• B. 50%
• C. 75%
• D. 100%

Answer 1

The correct option is A 33.33%

The balanced equation is:
$AgN{O}_3+NaCl\to AgCl+NaN{O}_3$

Moles of Silver Chloride = $\frac{169.8}{169.8}=1mol$
Moles of $NaCl$ = $\frac{29.2}{58.4}=0.5mol$

According to the balanced chemical equation, 1 mol of Silver Nitrate requires 1 mol of $NaCl$.
But the number of mol of $NaCl$ available is 0.5 mol.
Hence, $NaCl$ is the limiting reagent.

Therefore, the moles of Silver Chloride formed = 0.5 mol
Mass of Silver Chloride formed = $0.5\times 143.3$ = 71.6 g

Theoretical yield = 71.6 g
Experimental yield = 27.2 g

Percentage yield
=$\frac{\text{experimental yield}}{\text{theoretical yield}}\times 100=\frac{27.2}{71.6}\times 100=33.33\%$