Ratio of moles of Fe (II) oxidized by equal volumes of equimolar $K M n O_{4}$ and $K_{2} C r_{2} O_{7}$ solutions in acidic medium will be:

5 : 3

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1 : 1

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1 : 2

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5 : 6

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Solution

The correct option is D 5 : 6
$F e^{2 +} + M n O_{4}^{-} ⟶ F e^{3 +} + M n^{2 +}$
$n_{f} = 1$ $n_{f} = 5$
Moles of $F e^{+ 2} \times 1$ by $K M n O_{4} =$ moles of $M n O_{4}^{-} \times 5$ .....(i)
$F e^{2 +} + C r_{2} O_{7}^{2 -} \times ⟶ F e^{3 +} + C r^{3 +}$
$n_{f} = 1$ $n_{f} = 6$
Moles of $F e^{+ 2} \times 1$ by $K_{2} C r_{2} O_{7} =$ moles of $M n O_{4}^{-} \times 6$ .....(ii)
(i) $\div$ (ii)
$\frac{n_{1}}{n_{2}} = \frac{5}{6}$

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