Question

When a mixture of $NaBr$ and $NaCl$ is repeatedly digested with sulphuric acid, all the halogens are expelled and $Na_{2}SO_{4}$ is formed quantitatively. With a particular mixture, it was found that the weight of $Na_{2}SO_{4}$ 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

The reaction of sulphuric acid with $NaCl$ and $NaBr$ is given below:
$2NaCl+H_{2}SO_{4} \rightarrow Na_{2}SO_{4}+2HCl$
$2NaBr+H_{2}SO_{4} \rightarrow Na_{2}SO_{4}+2HBr$
The ratio between the total moles of $NaCl$ and $NaBr$ to the moles of $Na_{2}SO_{4}$ will be $2:1$ .
For making one mole of $Na_{2}SO_{4}$ , $2$ 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 $Na_{2}SO_{4}$ obtained is precisely the same as the mixture of $NaCl-NaBr$ taken.
$\therefore $ Grams $NaBr+$ Grams $NaCl=$ Grams $Na_{2}SO_{4}$

Moles of $NaCl=X=1.4348\,mol$
Moles of $NaBr=2-X=0.562\,mol$
$\therefore $ Weight of $NaCl=1.4348\times 58.5=84.13\,g$
Weight of $NaBr=0.562\times 103=57.886\,g$
Taking ratio,
$\frac{84 . 13}{57 . 886}=\frac{1 . 454}{1}$
Final answer $=100\times 1.45=145$