`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

`MnO_4^(ɵ)+5Fe^(2+)+8H^(o+)toMn^(2+)+5Fe^(3+)+4H_2O`
`47.2 " mL of " 0.112 N KMnO_4`
`=47.2xx0.112 m" Eq of "Fe^(2+)`
`=5.29xx10^(-3)xx56g of Fe^(+2)`
`=0.296g of Fe`
`% of Fe=(0.296xx100)/(0.804)=36.82%`
`3Fe=Fe_3O_4`
`3xx56=3xx56+64`
`therefore168 g of Fe=232 g of Fe_3O_4`
`%` of `Fe_2O_3=(232)/(168)xx36.8=50.85%`