Number of moles of KMnO4 required to oxidize one mole of FeC2O4 in acidic medium is

eq.  of  KMnO4=eq.  of  FeC2O4n×Valency factor=n×Valency factorx ×  5=1 × 3x=0.6  Note:  Mn               +7O4-+Fe+2+C+32O4-2→H+Mn+2+Fe+3+C+4O2 Change in oxidation state of Mn =valency factor =5 Change in oxidation state in  Fe(C2O4)=valency factor=1+2(1)=3