How many moles of $M n O_{4}^{⊝}$ ion will react with $1$ mol of ferrous oxalate in acidic medium?

\frac{1}{5}

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\frac{2}{5}

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\frac{3}{5}

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\frac{5}{3}

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Solution

The correct option is B $\frac{3}{5}$
$F e C_{2} O_{4} ⇌ F e^{2 +} + C_{2} O_{4}^{2 -}$
$M n O_{4}^{⊝}$ oxidises $F e^{2 +}$ and $C_{2} O_{4}^{2 -}$ both.
$F e^{2 +} + C_{2} O_{4}^{2 -} \to F e^{3 +} + 2 C O_{2} + 3 e^{-}$
$M n O_{4}^{⊝} + 5 e^{-} \to M n^{2 +}$
$5 F e C_{2} O_{4} \equiv 3 M n O_{4}^{⊝}$

$1 F e C_{2} O_{4} \equiv \frac{3}{5} M n O_{4}^{⊝}$

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