Question

$64g$ of a mixture of $NaCl$ and $KCl$ were treated with concentrated sulphuric acid. The total mass of metal sulphates obtained was found to be %$76g$. What are the mass percents of $NaCl$ in the mixture. The reactions are $2NaCl+H_{2}S{O}_4\to N{a}_{2}S{O}_4+2HCl$; $2KCl+H_{2}S{O}_4\to K_{2}S{O}_4+2HCl$;

• A. $42.89\%$
• B. $84.9\%$
• C. $31.5\%$
• D. $63.1\%$

Answer 1

The correct option is B $84.9\%$

Given,

Weight of $NaCl$and $KCl$mixture = 64 gm

Weight of metal sulphates, $N{a}_{2}S{O}_4$and $K_{2}S{O}_4$ = 76 gm

The reactions will be:

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

$2KCl+H_{2}S{O}_4\to K_{2}S{O}_4+2HCl$

Now, let mass of $NaCl$ be x gm.

Therefore, moles of $NaCl$ will be $=\frac{x}{58.5}$

And moles of $KCl$will be $=\frac{64-x}{74.5}$

From the above reactions we can write that

Moles of $NaCl$*1 = Moles of $N{a}_{2}S{O}_4$*2, or

Moles of $N{a}_{2}S{O}_4$ = Moles of $NaCl$* ½ $...\left(i\right)$

Similarly, for $KCl$ and $K_{2}S{O}_4$ we can write

Moles of $K_{2}S{O}_4$*1 = Moles of $KCl$* ½ $...\left(ii\right)$

It was given that the total weight of the metal sulphates was 76 gm, therefore from equation $\left(i\right)$and $\left(ii\right)$we get

Weight of $N{a}_{2}S{O}_4$ + Weight of $K_{2}S{O}_4$ = 76 gm

$\frac{1}{2}\times \frac{x}{58.5}\times 142+\frac{1}{2}\times \frac{64-x}{74.5}\times 174=76$

$\Rightarrow 1.21\times 74.4-1.16x=76$

Solving the above equation for $x$we get,

$x=27.45$

Now, percentage mass of $NaCl$ can be given as:

$=\frac{givenmass}{molarmass}\times 100$

$=\frac{27.45}{64}\times 100=42.89\%$

Hence, the mass percent of $NaCl$ in the mixture is $42.89\%$

Therefore, the correct option is A.