the ratio of numbers of mole of kmno4 and k2cr2o7 required to oxidised 0.1 mole sn2+ and sn4+ in acidic medium .
During the reaction KMnO4 and K2Cr2O7 are getting 5 and 6 electrons, respectively. At the same time Sn2+ loses 2 electrons. MnO4- + 5e + 8H+ →Mn2+ + 4H2O Cr2O72- + 6e + 14H+ →2Cr3+ + 7H2O Sn2+ - 2e → Sn+4 Thus, ratio of reagents should be KMnO4 : Sn2+ = 2 : 5 and K2Cr2O7 : Sn2+ = 1 : 3. Finally, the numbers of mole of KMnO4 and K2Cr2O7 are 0.04 mol and 0.033 mol required to oxidize 0.1 mole of Sn2+.