The weight of AgCl precipitated, when a solution containing 5.85 g of NaCl is added to a solution containing 3.4 g of AgNO3 is

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The weight of AgCl precipitated, when a solution containing 5.85 g of NaCl is added to a solution containing 3.4 g of AgNO₃ is

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28 g

9.25 g

2.87 g

58 g

answer is C.

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Detailed Solution

${AgNO}_{3} + NaCl \to AgCl + {NaNO}_{3}$
No. of moles of AgNO₃ $= \frac{3 .4}{170} = 0 .02$
No. of moles of NaCl $= \frac{5 .85}{58 .5} = 0 .1$
Limiting reagent = AgNO₃
1 mole of AgNO₃ produces 1 mole of AgCl
0.02 mole of AgNO₃ will produce 0.02 mole of AgCl
Weight of AgCl produced = 0.02 x 143.5 = 2.87 g