Question

Number of moles of $K_{2}C{r}_2{O}_7$ that can be reduced by 1 mol of $S{n}^{2+}$ ions is

• A. $\frac{1}{3}$
• B. $\frac{3}{2}$
• C. $\frac{5}{6}$
• D. $\frac{6}{5}$

Answer 1

Answer: (A)

$C{r}_2{O}_7^{2-}+14H^{+}+3S{n}^{2+}\to 2C{r}^{3+}+7H_{2}O+3S{n}^{4+}$
As 1 mole of $K_{2}C{r}_2{O}_7$ is oxidising 3 moles of $S{n}^{2+}$
So $1molS{n}^{2+}\equiv \frac{1}{3}molofC{r}_2{O}_7^{2-}$ .