0.7 g sample of iron ore was dissolved in acid. Iron was reduced to `+2` state and it required 50 " mL of " `(M)/(50)` `KMnO_(4)` solution for titration. The percentage of Fe and `Fe_(3)O_(4)` in the ore is

Options

A. `40%Fe,55.24%,Fe_(3)O_(4)`
B. `55.24%Fe,40%Fe_(3)O_(4)`
C. `8%Fe,11%Fe_(3)O_(4)`
D. `11%Fe,8%Fe_(3)O_(4)`

Answer

A

Explanation

Ore of iron is `Fe_(2)O_(4)` which is mixture of `FeO+Fe_(2)O_(3)`. In `Fe_(2)O_(3)` Fe is in `+3` state which is reduced to `+2` state
`(Fe^(3+)+e^(-)toFe^(2+))`
(In FeO, Fe is in `+2` state)
Fe in `+2` state (from FeO and `Fe_(2)O_(3)`) is equivalent of `MnO_(4)^(ɵ)`
`thereforeMnO_(4)^(ɵ)+5Fe^(2+)+H^(o+)toMn^(2+)+5Fe^(3+)+H_(2)O`
`m" Eq of "KMnO_(4)=50xx(1)/(50)xx5`
`=5mEq`
`=5 m" Eq of "Fe^(2+)`
`=5xx10^(-3)xx56g of Fe^(2+) or Fe`
`=0.280 g`
`% of Fe=(0.280)/(0.7)xx100=40%`
again `3Fe=Fe_(3)O_(4)`
`3xx56g-=(3xx56+64)g of Fe_(3)O_(4)`
`% of Fe_(3)O_(4)=(40xx232)/(168)=55.24%`