If equal volume of reactants are used, then no. of moles of `KMnO_(4)` (moles per litre) used in acidic modeium required to completely oxidise 0.5 M `FeSO_(3)`?

Options

A. 0.3
B. 0.1
C. 0.2
D. 0.4

Answer

A

Explanation

Both Fe(II) and S(IV) in `SO_(3)^(2-)` can be oxidised to Fe(III). And `(SO_(4))^(2-)` respectively hence `(3//5) xx0.5=0.3` mol/litre.
`[MnO_(4)^(-)+5e^(-)+8H^(+) to Mn^(2+)+4H_(2)O]xx(3)/(5) `
`Fe^(2+) to Fe^(3+)+e^(-)`
`SO_(3)^(2-) to SO_(4)^(2-)+2e^(-)`
`underlineoverline((3)/(5)MnO_(4)^(-)+(24)/(5)H^(+)+Fe^(2+)+SO_(3)^(2-) to 3//5Mn^(2+)+12//5H_(2)O+Fe^(3+)+SO_(4)^(2-))`