A 0.500 g sample contains only FeC2O4 (molar mass 143.86 g mol^-1, 60% by mass) and Fe2(C2O4)3 (molar mass 375.75 g mol^-1, 40% by mass). In acidic…

Use equivalents. For $\mathrm{FeC_2O_4}$, both $\mathrm{Fe^{2+}\to Fe^{3+}}$ ($1e^-$) and $\mathrm{C_2O_4^{2-}\to2CO_2}$ ($2e^-$) are oxidised, so $n$-factor $=3$. For $\mathrm{Fe_2(C_2O_4)_3}$, iron is already $\mathrm{Fe^{3+}}$ and only the three oxalate ions react, $n$-factor $=3\times2=6$. Moles: $\mathrm{FeC_2O_4}=0.300/143.86=2.0853\times10^{-3}$; $\mathrm{Fe_2(C_2O_4)_3}=0.200/375.75=5.3226\times10^{-4}$. Equivalents $=3(2.0853\times10^{-3})+6(5.3226\times10^{-4})=6.2559\times10^{-3}+3.1936\times10^{-3}=9.4495\times10^{-3}$. $\mathrm{KMnO_4}$ has $n$-factor $5$ so its normality is $0.100\ \mathrm{N}$. Volume $=9.4495\times10^{-3}/0.100=0.094495\ \mathrm{L}=94.50\ \mathrm{mL}$. The two $n$-factors differ because ferrous oxalate contributes iron as well as oxalate to the electron count, which is the key subtlety in this titration. If both compounds were wrongly given the same $n$-factor of $3$, the volume would come out near $78.53\ \mathrm{mL}$ instead.