A 4.0 g sample contained `Fe_2O_3,Fe_3O_4`, and inert material. It was treated with an excess of aq KI solution in acidic medium, which reduced all iron to `Fe^(2+)` ions. The resulting solution was diluted to 50 mL and a 10 mL sample of it was taken the iodine liberated in the small sample was titrated with 12.0 " mL of " 0.5 M `Na_2S_2O_3` solution. The iodine from another 25 mL was extracted, after which the `Fe^(2+)` ions were titrated with 16 " mL of " 0.25 M `MnO_4^(ɵ)` ions in `H_2SO_4` solution. Calculate the mass of two oxides in the original mixture.

Explanation

Let `Fe_2O_3-=xmmol`
`Fe_3O_4-=ymmol-=FeO+Fe_2O_3`
`Fe^(3+)=2(x+y)mmol and Fe^(2+)=y m mol`
`2Fe^(3+)+2I^(ɵ)toI_2+2Fe^(2+)`
`2m" mol of "Fe^(3+)-=1 mmol I_2`
`mmol I_2-=2m" mol of "S_2O_3^(2-)` (in 20 mL sample)
`(x+y)(10)/(50)=(1)/(2)xx12xx0.5`
`impliesx+y=15`
Note that now m" mol of "`Fe^(2+)` in 50 mL sample is
`2(x+y)+y=2x+3y`
Also,
`5Fe^(2+)+MnO_4^(ɵ)to5Fe^(3+)Mn^(2+)`
`5m" mol of "Fe^(2+)-=1 m" mol of "MnO_4^(ɵ)`
`implies(25)/(50)(2x+3y)=5(16xx0.25)`
`implies2x+3y=40`
Solving, we get `x=5 and y=10`
Mass of `Fe_2O_3=(5)/(1000)xx160=0.8g`
Mass of `Fe_2O_4=(10)/(1000)xx232=2.32g`