If equal volumes of 0.1 M $K M n O_{4}$ and $0.1 M K_{2} C r_{2} O_{7}$ solutions are allowed to oxidise $F e^{2 +}$ to $F e^{3 +}$ in acidic medium, then $F e^{2 +}$ oxidized will be:

Equal volumes of 1 M each of $K M n O_{4}$ and $K_{2} C r_{2} O_{7}$ are used to oxidise Fe(II) solution in acidic medium. The amount of Fe oxidized will be :

Q. 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 :

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If equal volumes of 1M KMnO4 and 1M K2Cr2O7 solutions are allowed to oxidize Fe2+ in acidic medium. The amount of iron oxidized will be:

The correct option is B more by K2Cr2O7 solutionM1V1n1=M2V2n2So for Kmno4n1=5 and for K2Cr2O7 is +3 in acidic.So, amount of Fe will be more by K2Cr2O7.

If equal volumes of $1 M K M n O_{4}$ and $1 M K_{2} C r_{2} O_{7}$ solutions are allowed to oxidize $F e^{2 +}$ in acidic medium. The amount of iron oxidized will be:

The correct option is B more by $K_{2} C r_{2} O_{7}$ solution
$\frac{M_{1} V_{1}}{n_{1}} = \frac{M_{2} V_{2}}{n^{2}}$
So for $K m n o_{4} n_{1} = 5$ and for $K_{2} C r_{2} O_{7}$ is $+ 3$ in acidic.
So, amount of $F_{e}$ will be more by $K_{2} C r_{2} O_{7}$ .