Correct Answer – 2
`underset(“Ferrous oxalate”)(Fe(COO)_(2)) rarr Fe^(2+)+(COO^(-))_(2)`
`underset(O.N=+2)(Fe^(2+)) overset(MnO_(2)^(+))(rarr) underset(O.N=+3)(Fe^(3+))`

In the first reaction, O.N changes by 1 unit while in the second reaction O.N changes by 2 units. Thus, total change in O.N is by 3 units. This implies the one mole of ferrous oxalate contains 3 equivalents.
`underset(Mn=+7)underset(O.N of)(MnO_(4)^(-))rarr underset(Mn=+2)underset(O.N. of)(Mn^(2+)) (“acidic medium”)`
Since in the above reaction, O.N. of Mn changes by 5 units, one mole of `MnO_(4)^(-)` contains 5 equivalents. According to the law of equivalence, we need 3 equivalents of `MnO_(4)^(-)` to oxidize one mole of ferrous oxalate completely in acidic medium. Since 5 equivalents of `MnO_(4)^(-)` come from 1 mole, 3 equivalents of `MnO_(4)^(-)` will come from
`=1/5xx3`
`=0.6 mol`

Correct Answer – A
`{:(Mn^(7+),+5e,rarr,Mn^(2+),),(,Fe^(2+),rarr,Fe^(3+),+e),(,(C^(3+))_(2),rarr,2C^(4+),+2e):}`
`:. 3` mole of `KMnO_(4) -= 5` mole of `FeC_(2)O_(4)`

Correct Answer – A
`3 underset(3M)MnO_(4)^(-) + 5 (Fe^(2+) + underset(5M)C_(2) O_(4)^(2-)) + 24 H^(+) rarr`
`3Mn_(2) + 5 Fe^(3+) + 10CO_(2) + 12 H_(2) O`
Thus `5M` of `Fe_(2)O_(4)` is oxidised by `3M` of `KMnO_(4)` then `1M` of `FeC_(2) O_(4)` is oxidised by `3//5` mole of `KMnO_(4)`

Correct Answer – a
In ferrous oxalate ferrous and oxalate ions are oxidized.
`Fe^(2+) rarr Fe^(3+)+e^(-)`
`C_(2)O_(4)^(2-) rarr 2CO_(2)+2e^(-)`
`MnO_(4)^(2-) rarr Mn^(2+)`
Thus eqn. will be `3MnO_(4)^(-)+5FeC_(2)O_(4) rarr`
`implies` 1 mole of ferrous oxalate requires `3//5` moles of `KMnO_(4)`, i.e., 0.6 mole.