Number of moles of $M n O_{4}^{-}$ required to oxidize 1 mole of ferrous oxalate completely in acidic medium will be?

0.4

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0.6

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0.2

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7.5

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Solution

The correct option is B 0.6
Let us take the reaction between ferrous oxalate and $M n O_{4}^{-}$ in presence of sulphuric acid as:

$F e C_{2} O_{4} + K M n O_{4} + H_{2} S O_{4} \to K_{2} S O_{4} + M n S O_{4} + F e_{2} (S O_{4})_{3} + H_{2} O + C O_{2}$

While, the balanced form of reaction is:

$10 F e C_{2} O_{4} + 6 K M n O_{4} + 24 H_{2} S O_{4} \to 3 K_{2} S O_{4} + 6 M n S O_{4} + 5 F e_{2} (S O_{4})_{3} + 24 H_{2} O + 20 C O_{2}$

So, we see that, 6 moles of $K M n O_{4}$ is required to oxidize 10 moles of $F e C_{2} O_{4}$ .

So, 1 mole of $F e C_{2} O_{4}$ would be oxidized by = 6/10 = 0.6 moles of $K M n O_{4}$ .
So, the answer is (a).

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