A 3.00 g sample containing `Fe_(3)O_(4), Fe_(2)O_(3)` and an inert impure substance, is treated with excess of KI solution in presence of dilute `H_(2)SO_(4)` . The entire iron is converted into `Fe^(2+)` along with the liberation of iodine. The resulting solution is diluted to 100 mL. A 20 mL of the dilute solution requires 11.0 mL of 0.5 M `Na_(2)S_(2)O_(3)` solution to reduce the iodine, present. A 50 mL of the dilute solution after complete extraction of the iodine required 12.80 mL of 0.25 M `KMnO_(4)` solution in dilute `H_(2)SO_(4)` medium for the oxidation of `Fe^(2+)` . Calculate the % of `Fe_(2)O_(3) "and" Fe_(3)O_(4)` in the original sample.

Explanation

The following reactions are involved in given data

`Fe_(2)O_(3) + 3H_(2)SO_(4) rarr Fe_(2)(SO_(4))_(3) + 3H_(2)O, FeO + H_(2)SO_(4) rarr FeSO_(4) + H_(2)O`
`Fe_(2)(SO_(4))_(3) + 2KI rarr 2FeSO_(4) + K_(2)SO_(4) + I_(2)`
`I_(2) + 2Na_(2)S_(2)O_(3) rarr 2 NaI + Na_(2)S_(4)O_(6)`
`2KMnO_(4) + 8H_(2)SO_(4) + 10FeSO_(4) rarr K_(2)SO_(4) + 2MnSO_(4) + 5Fe_(2)(SO_(4))_(3) + 8H_(2)O`
1 mole `Fe_(2)O_(3)` = 1 mole `Fe_(2)(SO_(4))_(3)` = 1 mole of `I_(2)` = 2 moles of `Na_(2)S_(2)O_(3)` and 5 moles of `FeSO_(4)` = 1 mole of `KMnO_(4)`
Number of millimoles of `I_(2)` in 20 mL = `5.5 xx 1/2 = 2.75 `
So, the number of millimoles of `I_(2)` in 100 mL = `2.75 xx 100/20 = 13.75`
Number of millimoles of `Fe_(2)O_(3) = 13.75`
Number of moles of `Fe_(2)O_(3) = 13.75 xx 10^(-3)`
Suppose x moles of `Fe_(3)O_(4)` and y moles of `Fe_(2)O_(3)` are present in 3g sample. In this sample `Fe_(3)O_(4)` is an equimolar mixture of FeO and `Fe_(2)O_(3)`. Therefore, total number of moles of `Fe_(2)O_(3)` in the mixture = x + y
`x + y = 13.75 xx 10^(-3)` ........ (i)
Number of millimoles of `KMnO_(4) = 0.25 xx 12.80 = 3.2`
Number of millimoles of `FeSO_(4)` in 50 mL = `3.2 xx 5 = 16` and number of millimoles of `FeSO_(4)` in 100 mL = `16 xx 100/50 = 32`

1 mole `Fe_(2)SO_(4)` gives 3 moles of `FeSO_(4)` and 1 mole of `Fe_(2)O_(3)` gives 2 moles of `FeSO_(4)`

`3x + 2y = 32 xx 10^(-3)` ........ (ii)
On solving equations (i) and (ii)
Number of moles of `Fe_(3)O_(4)` in the mixture `(x) = 4.5 xx 10^(-3)`
Number of moles of `Fe_(2)O_(3)` in the mixture `(y) = 9.25 xx 10^(-3)`
Mass of `Fe_(3)O_(4) = 4.5 xx 10^(-3) xx 232 = 1.044 g`
and Mass of `Fe_(2)O_(3) = 9.25 xx 10^(-3) xx 160 = 1.480 g`
` % of Fe_(3)O_(4) = 1.044/3.00 xx 100 = 34.8 %`
and `% of Fe_(2)O_(3) = 1.480/3.0 xx 100 = 49.33 %`