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(C) The oxidation of the components in an acidic medium involves the following $n$-factors:$FeSO_4 \rightarrow Fe^{3+} + SO_4^{2-} + e^-$,so $n$-facto

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$2$

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(C) The oxidation of the components in an acidic medium involves the following $n$-factors:$FeSO_4 ightarrow Fe^{3+} + SO_4^{2-} + e^-$,so $n$-factor $= 1$.$FeC_2O_4 ightarrow Fe^{3+} + 2CO_2 + 3e^-$,so $n$-factor $= 3$.$Fe_2(C_2O_4)_3 ightarrow 2Fe^{3+} + 6CO_2 + 6e^-$,so $n$-factor $= 6$.$Fe_2(SO_4)_3$ is already in the highest oxidation state $(Fe^{3+})$,so it does not react.For $KMnO_4$ in acidic medium,the $n$-factor is $5$.Using the principle of equivalence: $n_{eq}(KMnO_4) = n_{eq}(FeC_2O_4) + n_{eq}(Fe_2(C_2O_4)_3) + n_{eq}(FeSO_4)$.Let $x$ be the moles of $KMnO_4$: $x 	imes 5 = (1 	imes 3) + (1 	imes 6) + (1 	imes 1)$.$5x = 10$,which gives $x = 2$ moles.

In order to oxidize a mixture of one mole of each of $FeC_2O_4$,$Fe_2(C_2O_4)_3$,$FeSO_4$ and $Fe_2(SO_4)_3$ in acidic medium,the number of moles of $KMnO_4$ required is

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