Question

The number of moles of $K_2{Cr}_2{O}_7$ reduced by $1$ mol of ${Sn}^{2+}$ is:

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

Answer 1

Answer: (B)

$\frac{1}{3}$

Eq. of ${Cr}_2{O}_7^{2-}\equiv$ Eq. of ${Sn}^{2+}$
$(n=6)$ $(n=2)$
Therefore, $\frac{1}{6}$ mol $\equiv \frac{1}{2}$ mol
or, $\frac{1}{3}$ mol of ${Cr}_2{O}_7^{2-}\equiv$ 1 mol of ${Sn}^{2+}$