(i) A sample of `MnSO_(4).4H_(2)O ` is strongly heated in air . The residue is `Mn_(3)O_(4)`
(ii) The residue is dissolved in 100 mL of `0.1 ` N `FeSO_(4)` containing dilute `H_(2)SO_(4)`
(iii) The solution reacts completely with 50 mL of `KMnO_(4)` solution .
(iv) 25 mL of the `KMnO_(4)` solution used in step (ii) requires 30 mL of 0.1 N `FeSO_(4)` solution for complete reaction .

Explanation

m.e of 25 mL of `KMnO_(4)` solution
= m.e of 30 mL of 0.1 N `FeSO_(4)` solution
` = 0.1 xx 30 = 3`
` :. ` m.e of 50 mL of `KMnO_(4)` solution = `2 xx 3 = 6`
` :. ` m.e of `FeSO_(4)` (remained which did not react with `Mn_(3)O_(4)`) = 6
Now , m.e of total `FeSO_(4)` solution = `0.1 xx 100 = 10 `
`:. ` m.e of `Mn_(3)O_(4) = 4`
` :. ` eq. of `Mn_(3)O_(4) = 4/1000`
From the redox reaction
`{:(Mn_(3)O_(4)+Fe^(2+)to,3Mn^(2+)+Fe^(3+)),(+8,+6):}`
Equivalent wt. of `Mn_(3)O_(4) = (" mol .wt")/(" change in ON per mole") `
`= 229/2 = 114.50`
` :. " wt . of " Mn_(3)O_(4) = " equivalent " xx " eq.wt"`
` = 4/1000 xx 114.5 = 0.458 g `
As given in the problem `Mn_(3)O_(4)` is obtained by heating `MnSO_(4).4H_(2)O `
`MnSO_(4) . 4H_(2)O overset( Delta) to Mn_(3)O_(4)`
or `(" wt . of " MnSO_(4).4H_(2)O)/(" mol.wt.of "MnSO_(4).4H_(2)O) = (3xx 0.458)/229`
` :. " wt . of " MnSO_(4) .4H_(2)O = (3xx 0.458 )/229 xx 223 `
` = 1.338 g `