A 0.100 mol mixture of FeC2O4 and Fe2(C2O4)3 is titrated in acidic solution with KMnO4 (which is reduced to Mn^2+) and requires exactly 0.0900 mol of…

Work with equivalents. Each $\mathrm{MnO_4^-}$ gains $5e^-$, so $0.0900$ mol corresponds to $0.450$ equivalents. In $\mathrm{FeC_2O_4}$ both $\mathrm{Fe^{2+}\to Fe^{3+}}$ (1 electron) and $\mathrm{C_2O_4^{2-}\to2CO_2}$ (2 electrons) are oxidised, so its $n$-factor is $3$. In $\mathrm{Fe_2(C_2O_4)_3}$ iron is already $\mathrm{Fe^{3+}}$ and only the three oxalate ions react, so its $n$-factor is $3\times2=6$. Let the amounts be $a$ and $b$ mol: $a+b=0.100$ and $3a+6b=0.450$. Substituting $a=0.100-b$ gives $0.300+3b=0.450$, so $b=0.0500$ mol, equal to $5\times10^{-2}$ mol (and $a=0.0500$ mol). A common mistake is to give $\mathrm{Fe_2(C_2O_4)_3}$ an $n$-factor of $9$ by also oxidising iron which is not oxidisable. The answer is $5$.