Question

The weight of $AgCl$ precipitated when a solution containing 5.85 g of $NaCl$ is added to a solution containing 3.4 g of $AgN{O}_3$ is

• A. 28 g
• B. 9.25 g
• C. 2.87 g
• D. 58 g

Answer 1

Answer: (C)

$AgN{O}_3+NaCl\to AgCl+NaN{O}_3$
No. of moles of $AgN{O}_3=\frac{3.4}{170}=0.02$
No. of moles of $NaCl=\frac{5.85}{58.5}=0.1$
Limiting reagent $=AgN{O}_3$
$\because 1$ mole of $AgN{O}_3$ produces 1 mole of $AgCl$
$\therefore 0.02$ mole of $AgN{O}_3$ will produce $0.02$ mole of $AgCl$
Weight of $AgCl$ produced $=0.02\times 143.5$
$=2.870g$