Question

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

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

Answer 1

Answer: (A)

$\frac{1}{3}$

$\underset{(n=6)}{C{r}_2{O}_7^{2-}}=\underset{(n=2)}{S{n}^{2+}}(S{n}^{2+}\to S{n}^{4+}+2e^{-})$
$1$ Eq. of $C{r}_2{O}_7^{2-}$ = $1$ Eq. of $S{n}^{2+}$
$\frac{1}{6}$ mol$=\frac{1}{2}$ mol
$\therefore 1$ mol of $S{n}^{2+}=\frac{2}{6}$ mol of $C{r}_2{O}_7^{2-}$ $=\frac{1}{3}$ mol of $C{r}_2{O}_7^{2-}$.