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

${\mathrm{AgNO}}_3+\mathrm{NaCl}\to \mathrm{AgCl}+{\mathrm{NaNO}}_3$
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