Question

`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 1

The titration involves the conversion of ferrous into ferric.
`5Fe^(2+)+MnO_(4)^(-) +8H^(+) to 5Fe^(3+)+Mn^(2+)+4H_(2)O`
`47.2 mL " of" 0.112 N KMnO_(4)-=47.2 mL " of" 0.112 N Fe^(2+)`ions
`=(47.2xx0.112xx55.5)/(1000)=0.2934`
Mass of iron =0.2934 g
% of iron in the ore`=(0.2934)/(0.804)xx100=36.49`
`underset((3xx55.5))(3Fe) to underset((3xx55.5+64))(Fe_(3)O_(4))`
`3xx55.5` g of iron form 230.5 g of `Fe_(3)O_(4)`.
0.2934 g of iron will form `=(230.5)/(166.5)xx0.2934=0.406g`
% of `Fe_(3)O_(4)` in the ore `= (0.406)/(0.804)xx100=50.5`