The number of moles of $K_{2} {C r}_{2} O_{7}$ reduced by $3$ mol of ${S n}^{2 +}$ ions is :

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Solution

The redox changes are shown below:
$6 e^{-} + {C r}_{2} O_{7}^{2 -} ⟶ 2 {C r}^{3 +}$
${S n}^{2 +} ⟶ {S n}^{4 +} + 2 e^{-}$
Equivalent of ${C r}_{2} O_{7}^{2 -} =$ Equivalent of ${S n}^{2 +}$
$\Rightarrow 1$ Eq. $\equiv 1$ Eq.

$\frac{1}{6}$ mol $\equiv \frac{1}{2}$ mol

$1$ mol of ${C r}_{2} O_{7}^{2 -} = 3$ mol of ${S n}^{2 +}$

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