The number of moles of $Mn{\stackrel{\_}{O}}_4$  and $C{r}_2{O}_7^{2-}$ separately required to oxidize 1 mole of $Fe{C}_2{O}_4$ each in acidic medium respectively

$3Mn{\overline{O}}_4+5Fe{C}_2{O}_4+24H^{+}\to 3M{n}^{2+}+5F{e}^{3+}+10C{O}_2+12H_{2}O$

3 moles of $Mn{\overline{O}}_4$= 5 moles of $Fe{C}_2{O}_4$

                    ?          =1 moles of $Fe{C}_2{O}_4$

no. of moles of $Mn{\overline{O}}_4=\frac{3}{5}$

$C{r}_2{O}_7^{2-}+2Fe{C}_2{O}_4+14H^{+}\to 2C{r}_{3+}+2F{e}^{3+}+4C{O}_2+7H_{2}O$

2 moles of $Fe{C}_2{O}_4$ = 1 mole of $C{r}_2{O}_7^{2-}$

1 mole $Fe{C}_2{O}_4=?$

no of moles of $C{r}_2{O}_7^{2-}=\frac{1}{2}$