Question

Number of moles of $K_{2}Cr_{2}O_{7}$ that can be reduced by 1 mol of $Sn^{2 +}$ ions is

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

Answer 1

Answer: (A)

$Cr_{2}O_{7}^{2 -}+14H^{+}+3Sn^{2 +} \rightarrow 2Cr^{3 +}+7H_{2}O+3Sn^{4 +}$
As 1 mole of $K_{2}Cr_{2}O_{7}$ is oxidising 3 moles of $Sn^{2 +}$
So $1 \, mol \, Sn^{2 +}\equiv \frac{1}{3 \, }mol \, of \, Cr_{2}O_{7}^{2 -}$ .