A sample of `MnSO_4.4H_2O` is reacted in air to give `Mn_3O_4`. The residue `Mn_3O_4` is dissolved in 100 " mL of " `(N)/(12)FeSO_4` containing `H_2SO_4` The solution reacts completely with 50 " mL of " `KMnO_4`. 25 " mL of " this `KMnO_4` requires 30 " mL of " `(N)/(10)FeSO_4` for complete oxidation determine the amount of `MnSO_4.4H_2O` in the sample.

Explanation

`Mn_3O_4+3H_2SO_4to3MnSO_4+3H_2O+O`
`underline(2FeSO_4+H_2SO_4+OtoFe_2(SO_4)_3+H_2O)`
`underline(Mn_3O_4+4H_2SO_4+2FeSO_4to3MnSO_4+4H_2O+Fe_2(SO_4)_3)`
`Mn_3O_4+8H^(o+)+2Fe^(2+)to3Mn^(2+)+2F^(3+)+4H_2O`
25 " mL of " `KMnO_4` requires 30 " mL of " 0.1 N `Fe^(2+)`
`50mL` of `KMnO_4` requires 60 " mL of " 0.1 N `Fe^(2+)`
Volume of `FeSO_4` used by `Mn_3O_4`
`=100-60=40" mL of " 0.1 N Fe^(+2)`
`=40xx0.1 m" Eq of "FeSO_4`
`=4m" mol of "FeSO_4`
`6 m" mol of "MnSO_4`
`=6xx10^(-3)xx223g of MnSO_4`
`=1.338 g of MnSO_4`
[2 m" mol of "`FeSO_4=1 m" mol of "Mn_3O_4)
`=3 m" mol of "MnSO_4]`