One mole of ferrous oxalate is oxidized by $x$ mole of $M n O_{4}^{-}$ in acidic medium. $x$ is:

0.6

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0.1

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0.3

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1.0

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Solution

The correct option is A $0.6$

$10 F e C_{2} O_{4} + 6 K M n O_{4} + 24 H_{2} S O_{4} ⟶ 3 K_{2} S O_{4} + 6 M n S O_{4} + 5 F e_{2} (S O_{4})_{3} + 24 H_{2} O + + 20 C O_{2}$
So we see that 6 moles of $K M n O_{4}$ is required to oxidize 10 moles of $F e C_{2} O_{4}$
Then, 1 mole of $F e C_{2} O_{4}$ would be oxidized by = ?
Cross multiply , so we get $\frac{6}{10} = 0.6$

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