Question

Number of moles of $K_{2}C{r}_2{O}_7$ reduced by one mole of $S{n}^{2+}$ ions is :

• A. $\frac{1}{3}$
• B. $3$
• C. $\frac{1}{6}$
• D. $6$

Answer 1

The correct option is A $\frac{1}{3}$

Number of moles of $K_{2}C{r}_2{O}_7$ reduced by one mole of $S{n}^{2+}$ ions is $\frac{1}{3}$. The balanced reaction is given below:
$C{r}_2{O}_7^{2-}+14H^{+}+3S{n}^{2+}\to 2C{r}^{3+}+7H_{2}O+3S{n}^{4+}$
From the reaction, we can see $3$ moles of $S{n}^{2+}$ is oxidized by $1$ mole of $C{r}_2{O}_7^{2-}$.
Hence, $1$ mole of $S{n}^{2+}$ is oxidized by $\frac{1}{3}$ mole of $C{r}_2{O}_7^{2-}$.