Question

The number of moles of $NaCl$ required to react with an excess of $AgN{O}_3$ to produce $14.34\text{g}$ of $AgCl$ is:

(Molar mass of $AgCl$ is $143.43{\text{g mol}}^{-1}$)

• A. $0.1$ moles
• B. $0.2$ moles
• C. $0.05$ moles
• D. $0.025$ moles

Answer 1

The correct option is A $0.1$ moles

$AgN{O}_3+NaCl\stackrel{}{\to }AgCl+NaN{O}_3$

Molar Mass of $AgCl$ = $143.32\text{g}$

Number of moles of $AgCl$ in $14.34\text{g}$ of $AgCl$ = $\frac{\text{Given mass of}AgCl}{\text{Molar Mass of}AgCl}$
$=\frac{14.34}{143.32}=0.1\text{mol}$

1 mole of $NaCl$ produces 1 mole of $AgCl$

So, the number of moles of $NaCl$ that produces $0.1\text{mol}$ of $AgCl$
= $0.1\text{mol}$