A 6.00g sample contained `Fe_(3) O_(4), Fe_(2) O_(3)` and inert materials. It was treated with an excess of aqueous Kl in acidic medium which reduced all the iron to `Fe^(2+)`. The resulting solution was dilution to 50ml. and a 10ml. sample of it was taken. The liberated iodine reacts with 5.5ml. of 1M `Na_(2) S_(2) O_(3)` solution, yielding `S_(4) O_(6)^(2-)`. The iodine from another 25ml. sample was extracted, after which the `Fe^(2+)` was titrated with 3.2 ml. of 1M `MnO_(4)^(-)` in `H_(2)SO_(4)` solution. Calculate the percentage of `Fe_(3) O_(4)` and `Fe_(2)O_(3)` in the original mixture.

Explanation

Equivalents of `Na_(2)S_(2) O_(3)` reacting with the liberated `I_(2) = ( 5.5 xx 1)/( 1000) = 5.5 xx 10^(-3)`
`:.` Equivalent of `I_(2)` in 10ml `= 5.5 xx 10^(-3)`
Equivalent of `Fe^(3+)` in 50ml = `5.5 xx 10^(-3) xx 5 = 0.0275`[ lt is `Fe^(3+)` that reacts with Kl to give `I_(2) `]
Total moles of `Fe_(2) O_(3) = ( 0.0275)/( 2) = .01375` ( free `+` combined )
Equivalengt of `MnO_(4)^(-) = ( 1 xx 5 xx 3.2 )/( 1000) = 0.016`
`:.` Eq of `Fe^(2+)` in 25 ml = 0.016
Eq of `Fe^(2+)` in 50ml = `0.016 xx 2 = 0.0 32 `
Eq. of FeO `= 0.032 - 0.0275 = 4.5 xx 10^(-3)`
Moles of `FeO = 4.5 xx 10^(-3)`
`:.` Moles of `Fe_(3) O_(4) = 4.5 xx 10^(-3)`
Mass of `Fe_(3) O_(4) = 4.5 xx 10^(-3) xx 232 = 1.044g m`.
Moles of `Fe_(2) O_(3) ` present freely `= 0.01375 - 4.5 xx 10^(-3) = 9.25 xx 10^(-3)`
Mass of `Fe_(2) O_(3) ` present freely `= 9.25 xx 10^(-3) xx 160 = 1.48 g `
% of `Fe_(3) O_(4) = ( 1.044)/(6) xx 100 = 17.4 % `
% `Fe_(2) O_(3) = 24.67%`