• # question_answer In order to oxidise a mixture of one mole of each of $Fe{{C}_{2}}{{O}_{4}},$ $F{{e}_{2}}({{C}_{2}}{{O}_{4}})3,$ $FeS{{O}_{4}}$ and $F{{e}_{2}}{{(S{{O}_{4}})}_{3}}$ in acidic medium, the number of moles of $KMn{{O}_{4}}$required is : A) 2                      B) 1 C) 1.5                                D) 3

 ${{F}_{e}}{{C}_{2}}{{O}_{4}}\to F{{e}^{+3}}+C{{O}_{2}}$ $F{{e}_{2}}{{({{C}_{2}}{{O}_{4}})}_{3}}\to C{{O}_{2}}$ $F{{e}_{2}}{{(S{{O}_{4}})}_{3}}\to \,No\,oxidation$ $FeS{{O}_{4}}\to F{{e}^{+3}}$ ${{g}_{m}}E\,(KMn{{O}_{4}})={{g}_{m}}E\,(Fe{{C}_{2}}{{O}_{4}})+{{g}_{m}}E\,\,[F{{e}_{2}}{{({{C}_{2}}{{O}_{4}})}_{3}}]+{{g}_{m}}E\,(FeS{{O}_{4}})$ $moles\,\times \,V.F=moles\times V.F.+moles\times V.F.+moles\times V.F.$ $x\times 5=(1\times 3)+(1\times 6)+(1\times 1)$ $5x=10\,(MnO_{4}^{-}\to M{{n}^{+2}})$ $x=2.$