The ratio of the number of moles of $K M n O_{4}$ and $K_{2} C r_{2} O_{7}$ required to oxidise 0.1 mol $S n^{2 +}$ to $S n^{+ 4}$ in acidic medium _____.

6 : 5

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5 : 6

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1 : 2

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2 : 1

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Solution

The correct option is A 6 : 5
$2 {K M n O}_{4} + 5 {S n}^{2 +} + 16 H^{+} \to {5 S n}^{4 +} + 2 {M n}^{2 +} + 8 H_{2} O + 2 K^{+}$
${(0.1 m o l {S n}^{2 +}) \times (2 m o l {K M n O}_{4} / 5 m o l {S n}^{2 +})} = 0.04 m o l {K M n O}_{4}$
$K_{2} {C r}_{2} O_{7} + 3 {S n}^{2 +} + 14 H^{+} \to 3 {S n}^{4 +} + 2 {C r}^{3 +} + 7 H_{2} O + 2 K^{+}$
${(0.1 m o l {S n}^{2 +}) \times (1 m o l K_{2} {C r}_{2} O_{7} / 3 m o l {S n}^{2 +})} = 0.0333 m o l K_{2} {C r}_{2} O_{7}$
$(0.04 m o l {K M n O}_{4}) / (0.0333 m o l K_{2} {C r}_{2} O_{7}) = \frac{0.12}{0.1} = 1.2 = \frac{6}{5}$
So, the ratio required to oxidize is $6 : 5$

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