Number of moles of `MnO_(4)^(-)` required to oxidise one mole of ferrous oxalate completely in acidic medium will be

Answer

`0.6`

Explanation

`MnO_(4)^(-)+8H^(+)+5e^(-)rarrMn^(2+)+4H_(2)O`
`FeC_(2)O_(4)rarrFe^(2+)+C_(2)O_(4)^(2-)`
`[(Fe^(2+)rarrFe^(3+)+e^(-)),(C_(2)O_(4)^(2-)rarr2CO_(2)+2e^(-))]`
Total no. of electrons involved = 3
Since one mole of `FeC_(2)O_(4)` loses 3 moles of electrons while one mole of `KMnO_(4)` accepts five moles of electrons, therefore, number of moles of `KMnO_(4)` required to oxidise one mole of `FeC_(2)O_(4)=3//5=0.6` mole.