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 ${AgNO_3}$ is

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

Answer 1

Answer: (C)

2.87 g

Answer 2

${AgNO_3 + NaCl ? AgCl + NaNO_3}$ No. of moles of ${AgNO_3 = \frac{3.4}{170} = 0.02}$ No. of moles of ${NaCl = \frac{5.85}{58.5} = 0.1}$ Limiting reagent =${ AgNO_3}$ $\because$ 1 mole of ${AgNO_3}$ produces 1 mole of ${AgCl}$ $\therefore$ 0.02 mole of ${AgNO_3}$ will produce 0.02 mole of ${AgCl}$ Weight of ${AgCl}$ produced = $0.02 \times 143.5$ = $2.870 \, g$