`0.804 g` sample of iron ore was dissolved in acid. Iron was oxidised to `+2` state and it requires `47.2 mL` of `0.112N KMnO_(4)` solution for titration, Calculate `%` of `Fe` and `Fe_(3)O_(4)` in ore.

Explanation

`"Meq.of" Fe^(2+)="Meq.of" KMnO_(4)`
`"Meq.of" Fe^(2+)=47.2xx0.112=5.2864`
`:. (w)/(56//2)xx1000=5.2864`
`because Fe^(2+)rarrFe^(3+)+1e`
`therefore w_(Fe^(2+))=0.296 g`
`therefore %` Purity of `Fe=(0.296xx100)/(0.804)=36.82%`
Now `Fe_(3)O_(4)rarr3Fe`
`3xx56gFe` is obtained from `232 g Fe_(3)O_(4)`.
`therefore % "of" Fe_(3)O_(4)=(0.409)/(0.804)xx100=50.87%`