`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.

`0.804 g` sample of iron ore was dissolved in acid. Iron was oxidised

1.	`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.
Answer» `"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%`

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`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.

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