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In order to o...

Question

In order to oxidise a mixture of one mole of each of $F e C_{2} O_{4}, F e_{2} (C_{2} O_{4})_{3}, F e S O_{4} and F e_{2} (S O_{4})_{3}$ in acidic medium, the number of moles of $K M n O_{4}$ required is :

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1.5

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Solution

The correct option is A 2
Change in oxidation state for respective species:

$F e^{2 +} \to F e^{3 +} + e^{-}$

$C_{2} O_{4}^{2 -} \to 2 C O_{2} + 2 e^{-}$

$M n O_{4}^{-} \to M n^{2 +} + 5 e^{-}$

There will be no change in the oxidation state of $F e_{2} (S O_{4})_{3}$ as $F e$ is already in the highest oxidation state.

Using equivalence concept:

$n_{e q} (K M n O_{4}) = n_{e q} [F e C_{2} O_{4}] + n_{e q} [F e_{2} (C_{2} O_{4})_{3}] + n_{e q} [F e S O_{4}]$
Let $n$ be the number of moles of $K M n O_{4}$
$n \times 5 = (1 \times 3) + (1 \times 6) + (1 \times 1) 5 n = 10$

$∴ n = 2$

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