The number of moles of $K M n O_{4}$ that will be needed to react completely with 1 mole of ferrous oxalate in acidic medium is

0.6

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0.600

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0.60

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Solution

The balanced reaction is
$3 M n O_{4}^{-} + 5 F e C_{2} O_{4} + 24 H^{+} \to 3 M n^{2 +} + 5 F e 3 + + 10 C O_{2} + 12 H_{2} O$

$\Rightarrow 5 m o l o f F e C_{2} O_{4}$ requires $3 m o l o f K M n O_{4}$

$\Rightarrow 1 m o l o f F e C_{2} O_{4}$ will require $\frac{3}{5} m o l o f K M n O_{4}$

i.e., $0.6$ $m o l o f K M n O_{4}$

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