Question

When a mixture of $NaBr$ and $NaCl$ is repeatedly digested with sulphuric acid, all the halogens are expelled and $N{a}_{2}S{O}_4$ is formed quantitatively. With a particular mixture, it was found that the weight of $N{a}_{2}S{O}_4$ obtained was precisely the same as the weight of $NaBr-NaCl$ mixture taken. Calculate the ratio of the weights of $NaCl$ and $NaBr$ in the mixture.
(Note: Report your answer after multiplying with $100$ and rounding up to the nearest integer value.)

Answer 1

Answer: 145

The reaction of sulphuric acid with $NaCl$ and $NaBr$ is given below:
$2NaCl+H_{2}S{O}_4\to N{a}_{2}S{O}_4+2HCl$
$2NaBr+H_{2}S{O}_4\to N{a}_{2}S{O}_4+2HBr$
The ratio between the total moles of $NaCl$ and $NaBr$ to the moles of $N{a}_{2}S{O}_4$ will be $2:1$ .
For making one mole of $N{a}_{2}S{O}_4$ , $2$ moles of $NaCl+NaBr$ are required.
Let the moles of $NaCl=X$ , so the moles of $NaBr=2-X$ .
It is given that the weight of $N{a}_{2}S{O}_4$ obtained is precisely the same as the mixture of $NaCl-NaBr$ taken.
$\therefore$ Grams $NaBr+$ Grams $NaCl=$ Grams $N{a}_{2}S{O}_4$

Moles of $NaCl=X=1.4348mol$
Moles of $NaBr=2-X=0.562mol$
$\therefore$ Weight of $NaCl=1.4348\times 58.5=84.13g$
Weight of $NaBr=0.562\times 103=57.886g$
Taking ratio,
$\frac{84.13}{57.886}=\frac{1.454}{1}$
Final answer $=100\times 1.45=145$