A sample of `MnSO_(4). 4H_(2)O` is strongly heated in air. The residue `( Mn_(3)O_(4))` left was dissolved in `100 mL` of `0.1NFeSO_94)` containing dil. `H_(2)SO_(4)`. This solution was completely reacted with `50 mL` of `KMnO_(4)` solution. `25 mL` of this `KMnO_(4)` solution was completely reduced by `30 mL` of `0.1N FeSO_(4)` solution. Calculate the amount of `MnSO_(4).4H_(2)O` in sample.

Explanation

`MnSO_(4).4H_(2)OrarrMn_(3)O_(4)`
`3Mn^(2+)rarrMn_(3)^((8//3))+2e`
The residue `Mn_(3)O_(4)` is dissolved in `FeSO_(4)` which is reduced from `Mn^((8//3+))"to" Mn^(2+).Mn_(3)^((8//3+))+2erarr3Mn^(2+)`. The excess of `FeSO_(4)` is titrated by `KMnO_(4)`. The normality of `KMnO_(4)` is determined by another `FeSO_(4)`. For normality of `KMnO_(4)`:
Meq.of `KMnO_(4)="Meq.of" FeSO_(4)`
`25xxN=30xx0.1`
`therefore N=(3)/(25)`
Now Meq.of `FeSO_(4)` left after reacation with `Mn_(3)O_(4)= "Meq.of" KMnO_(4)` used `=50xx(3)/(25)=6`
`:. "Meq.of" FeSO_(4)"used for" Mn_(3)O_(4)=10-6=4`
`:. "Meq.of" MnSO_(4).4H_(2)O=4`
`because "For" MnSO_(4), E=(M)/(2//3)=(3M)/(2)`
`:. (w)/(3M//2)xx1000=4`
`:. (wxx2)/(3xx223)xx1000=4`