$M n O_{4}^{-}$ is an oxidant for $F e C_{2} O_{4}$ and other complexes. Mention the moles of $M n O_{4}^{-}$ required to oxidise $1$ mol of ferrous oxalate.

0.6 m o l

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3.6 m o l

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2.6 m o l

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1.6 m o l

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Solution

The correct option is B $0.6 m o l$
The reactions are as follows:
$F e C_{2} O_{4} ⇌ {F e}^{2 +} + C_{2} O_{4}^{2 -}$
$5 {F e}^{2 +} + M n O_{4}^{-} ⟶ {M n}^{2 +} + 5 {F e}^{3 +}$
$5 C_{2} O_{4}^{2 -} + 2 M n O_{4}^{-} ⟶ 2 {M n}^{2 +} + 10 {C O}_{2}$
$\Rightarrow$ 5 moles of $F e C_{2} O_{4}$ requires 3 moles of $M n O_{4}^{-}$
$∴$ 1 mol of $F e C_{2} O_{4}$ requires 0.6 mol of $M n O_{4}^{-}$
Therefore, moles of $M n O_{4}^{-}$ required by $1$ mol of ferrous oxalate is $0.6$ mol.

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