Question

What weight of silver chlorides will be precipitated when a solution containing $4.68gm$ of $NaCl$ is added to a solution of $5.10gm$ of $AgN{O}_3$?

• A. $4.37gm$
• B. $4.31gm$
• C. $5.97gm$
• D. $3.31gm$

Answer 1

The correct option is B $4.31gm$

According to the equation,
$NaCl+AgN{O}_3\to NaN{O}_3+AgCl$
Number of moles of $NaCl=\frac{4.68}{58.5}=0.08$
Number of moles of $AgN{O}_3=\frac{5.10}{170}=0.03$
Thus $AgN{O}_3$ is the limiting reagent in this reaction.
From stoichiometry of the reaction,
1 mole of $AgN{O}_3$ produces 1 mole of $AgCl$.
$\therefore$ $0.03$ mole of $AgN{O}_3$, produces $0.03$ mole of $AgCl$
$\therefore$ weight of precipitated $AgCl$ = $0.03\times \text{molar mass of}AgCl$
$=0.03\times 143.5$
$\approx 4.31gm$