Question

64 g of a mixture of NaCl and KCl were treated with concentrated sulphuric acid. The total mass of the metal sulphates obtained was found to be 76 g. What is the mass percent of NaCl in the mixture?

(Molar mass of KCl =74.5$gmo{l}^{-1}$,
NaCl = 58.5 $gmo{l}^{-1}$,
Molar mass of $N{a}_{2}S{O}_4$ =142 $gmo{l}^{-1}$ and Molar mass of $K_{2}S{O}_4$= 174 $gmo{l}^{-1}$

$NaCl+H_{2}S{O}_4\to N{a}_{2}S{O}_4+HClKCl+H_{2}S{O}_4\to K_{2}S{O}_4+HCl$

• A. 42.85%
• B. 84.91%
• C. 31.50%
• D. 63.17%

Answer 1

The correct option is A 42.85%

Consider the mass of NaCl to be x g.
Moles of NaCl = $\frac{x}{58.5}$ and moles of KCl = $\frac{64-x}{74.5}$
Applying the principle of atom conservation (POAC) to Na,
Moles of sodium in NaCl = moles of sodium in $N{a}_{2}S{O}_4$
1 $\times$moles of NaCl = 2 $\times$moles of $N{a}_{2}S{O}_4$
Moles of $N{a}_{2}S{O}_4$ = $\frac{1}{2}\times \text{moles of NaCl}$
Applying the principle of atom conservation (POAC) to K,
Moles of potassium in KCl = moles of potassium in $K_{2}S{O}_4$
1 $\times$moles of KCl = 2 $\times$ moles of $K_{2}S{O}_4$
Moles of $K_{2}S{O}_4$ = $\frac{1}{2}\times \text{moles of KCl}$
Total weight of $N{a}_{2}S{O}_4$ and $K_{2}S{O}_4$ is 76 g
Hence,
$\frac{1}{2}\times \frac{x}{58.5}\times 142+\frac{1}{2}\times \frac{64-x}{74.5}\times 174=76$
1.214x + 74.738 - 1.168x = 76
0.046x = 1.262
x = 27.43 g
% mass of NaCl = $\frac{27.43}{64}\times 100=42.85\%$