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(A) The reduction half-reaction for $KMnO_4$ in an acidic medium is:$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$Thus,$1$ mole of $KMnO_4$ provi

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$2$

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(A) The reduction half-reaction for $KMnO_4$ in an acidic medium is:$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$Thus,$1$ mole of $KMnO_4$ provides $5$ moles of electrons.Now,calculate the moles of electrons released by each compound:$1$. For $FeC_2O_4$: $Fe^{2+} \rightarrow Fe^{3+} + e^-$ and $C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-$. Total electrons = $1 + 2 = 3$ moles.$2$. For $Fe_2(C_2O_4)_3$: $2Fe^{2+} \rightarrow 2Fe^{3+} + 2e^-$ and $3C_2O_4^{2-} \rightarrow 6CO_2 + 6e^-$. Total electrons = $2 + 6 = 8$ moles.$3$. For $FeSO_4$: $Fe^{2+} \rightarrow Fe^{3+} + e^-$. Total electrons = $1$ mole.$4$. For $Fe_2(SO_4)_3$: Iron is already in the $+3$ oxidation state,so it cannot be oxidized further. Electrons = $0$ moles.Total moles of electrons = $3 + 8 + 1 = 12$ moles.Since $1$ mole of $KMnO_4$ accepts $5$ moles of electrons,the moles of $KMnO_4$ required = $\frac{12}{5} = 2.4$ moles.Wait,re-evaluating the standard stoichiometry for $Fe_2(C_2O_4)_3$:$Fe_2(C_2O_4)_3$ contains $Fe^{3+}$ and $C_2O_4^{2-}$. Only the oxalate part is oxidized: $3C_2O_4^{2-} \rightarrow 6CO_2 + 6e^-$.Total electrons = $3 (FeC_2O_4) + 6 (Fe_2(C_2O_4)_3) + 1 (FeSO_4) = 10$ moles.Moles of $KMnO_4 = \frac{10}{5} = 2$.

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 an acidic medium,the number of moles of $KMnO_4$ required is:

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