Question

One line of an acidified solution of $KMn{O}_4$ containing $15$ g of $KMn{O}_4$ is decolourised by passing sufficient amount of $S{O}_2$. If $S{O}_2$ is produced by roasting of iron pyrites $\left(Fe{S}_2\right)$ and $X$ g amount of pyrites is required to produce the necessary amount of $S{O}_2$, then what is the value of $X$?

• A. $15$
• B. $20$
• C. $10$
• D. None of these

Answer 1

The correct option is A $15$

The reaction for the roasting of $Fe{S}_2$ is as follows:
$\underset{4\times 120}{4Fe{S}_2}+11O_2\to 2F{e}_2{O}_3+\underset{8\times 64}{8S{O}_2}$
Equivalents of $KMn{O}_4=\frac{15.0}{31.6}\times 1$ L$=0.5$ eq. of $KMn{O}_4$ $=0.5$eq of $S{O}_2$ $=0.5\times \frac{64}{2}g$ of $S{O}_2$ $=16$ g of $S{O}_2$
$8\times 64$ g of $S{O}_2$ is produced from $2\times 120$ g of $Fe{S}_2$. (from the equation above)
$\therefore 16$ g of $S{O}_2$ is produced from $\frac{4\times 120}{8\times 64}\times 64=15$ g.