The number of moles of $M n O_{4}^{-}$ and $C r_{2} O_{7}^{2 -}$ separately required to oxidize 1 mole of $F e C_{2} O_{4}$ each in acidic medium respectively are:

0.6, 0.5

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0.6, 0.4

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0.5, 0.6

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0.4, 0.5

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Solution

The correct option is A 0.6, 0.5
$3 M n O_{4}^{-} + 5 F e C_{2} O_{4} + 24 H^{+} ⟶ 3 M n^{+ 2} + 5 F e^{+ 3} + 10 C O_{2} + 12 H_{2} O$

$⟹ 5 F e C_{2} O_{4}$ reacts with $3 M n O_{4}^{-}$

$⟹ 1$ $m o l e$ of $F e C_{2} O_{4} = \frac{3}{5}$ $m o l e$ of $M n O_{4}^{-}$

$∴ 1$ $m o l e$ of $F e C_{2} O_{4} = 0.6$ $m o l e$ of $M n O_{4}^{-}$

Again,

$C r_{2} O_{7}^{2 -} + 2 F e C_{2} O_{4} + 14 H^{+} ⟶ 2 C r^{+ 3} + 2 F e^{+ 3} + 4 C O_{2} + 7 H_{2} O$

$⟹ 2$ $m o l e$ of $F e C_{2} O_{4}$ required $1$ $m o l e$ of $C r_{2} O_{7}^{2 -}$

$⟹ 1$ $m o l e$ of $F e C_{2} O_{4}$ required $\frac{1}{2}$ $m o l e$ of $C r_{2} O_{7}^{2 -}$

$⟹ 0.5$ $m o l e$ of $C r_{2} O_{7}^{2 -}$

$1$ $m o l e$ of $F e C_{2} O_{4}$ required $= 0.6$ $m o l e$ of $M n O_{4}^{-}$ and $0.5$ $m o l e$ of $C r_{2} O_{7}^{2 -}$

Hence, the correct option is A

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