The number of moles of $$KMn{O_4}$$  that are needed to react completely with one mole of ferrous oxalate in acidic solution is

Solution :
$$\eqalign{ & F{e^{2 + }} + {C_2}O_4^{2 - } \to F{e^{3 + }} + 2C{O_2} + 3{e^ - } \cr & MnO_4^ - + 5{e^ - } \to M{n^{2 + }} \cr} $$
$$1\,mole$$  of $$KMn{O_4}$$  accepts 5 electrons
$$1\,mole$$  of ferrous oxalate loses 3 electrons
$$5{e^ - } \equiv 1\,mole\,\,{\text{of}}\,\,KMn{O_4}$$
$$3{e^ - }$$  will be equivalent to $$\frac{3}{5}\,mole$$  of $$KMn{O_4}$$