500 mid of 0.25 M $N{a}_{2}S{O}_4$ solution is added to an aqueous solution of 15 g of $BaC{l}_2$ resulting in the formation of a white precipitate of insoluble $BaS{O}_4$. How many moles and how many grams of $BaS{O}_4$ are formed?

Let the total amount of mixture of $NaCl$ and $NaI$.

taken initially was $x$gm. out of which amount of $NaCl$ was $y$ gm.

So,

Amount of $NaI$ in the initial mixture was $(x-y)$ gm.

$\begin{array}{c}2NaCl+H_{2}S{O}_4\to N{a}_{2}S{O}_4+HCl\\ 2NaI+H_{2}S{O}_4\to N{a}_{2}S{O}_4+HI\end{array}$

Amount of reacting psychometric.

$\begin{array}{c}NaClN{a}_{2}S{O}_{4}NaIN{a}_{2}S{O}_4\\ 117142300142\\ y\frac{142\times y}{117}x-y\frac{142\left(x-y\right)}{300}\end{array}$

at equilibrium,

$\begin{array}{c}\frac{142\times y}{117}+\frac{142\left(x-y\right)}{300}=x\\ \Rightarrow 14200y+553x-55384=1170\\ \Rightarrow 8662y=6162x\\ \left(\frac{y}{x}\right)\times 100=\left(\frac{6162}{8662}\right)\times 100=71.14\end{array}$

This is the percrnt of $NaCl$ present in the mixture.

Therefore,percent of $NaI$ present in the original mixture was

$100-71.14$

$=23.86$