Question

10 g NaCl solution is mixed with 17 g of silver nitrate solution. Calculate the weight of silver chloride precipitated.
AgNO₃ + NaCl $\stackrel{}{\to }$ AgCl + NaNO₃ [Ag = 108]

Answer 1

We know that,
Molecular mass of NaCl = 58.5 g
Molecular mass of AgNO₃ = [108 + 14 + 48] g = 170 g
Number of moles of NaCl = $\frac{\mathrm{Mass}\mathrm{of}\mathrm{NaCl}}{\mathrm{Molecular}\mathrm{mass}\mathrm{of}\mathrm{NaCl}}=\frac{10}{58.5}=0.17$

Number of moles of AgNO₃ = $\frac{\mathrm{Mass}\mathrm{of}{\mathrm{AgNO}}_3}{\mathrm{Molecular}\mathrm{mass}\mathrm{of}{\mathrm{AgNO}}_3}=\frac{17}{170}=0.1$
According to the balanced chemical equation, one mole of AgNO₃ reacts completely with one mole of NaCl.
So, 0.1 moles of AgNO₃ will react completely with only 0.1 moles of NaCl. Thus, AgNO₃ is the limiting reagent. The amount of product AgCl formed will depend upon the amount of AgNO₃ available for reaction.
As per the given reaction,
Moles of AgCl produced by one mole of AgNO₃ = 1
Moles of AgCl produced by 0.1 mole of AgNO₃ = 0.1
Mass of 0.1 mole of AgCl = Number of moles $\times$ Molecular mass of AgCl
= (0.1 $\times$ 143.5) g = 14.35 g