How many moles of $K M n O_{4}$ are reduced by $1$ mole of ferrous oxalte in acidic medium?

\frac{1}{5}

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\frac{5}{3}

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\frac{1}{3}

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\frac{3}{5}

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Solution

The correct option is D $\frac{3}{5}$
The balanced chemical equation of the reaction is-

$3 M n O_{4}^{-} + 5 F e C_{2} O_{4} + 24 H^{+} ⟶ 3 M n^{2 +} + 5 F e^{3 +} + 10 C O_{2} + 12 H_{2} O$

Therefore $3$ mole of $K M n O_{4}$ will be required

$3$ mole of $K M n O_{4} = 5$ moles of $F e C_{2} O_{4}$

$\frac{3}{5}$ mole of $K M n O_{4} =$ 1 moles of $F e C_{2} o_{4}$

For 1 mole $F e C_{2} O_{4}, \frac{3}{5} K M N O_{4}$ is required

So, the correct option is $D$

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