A 8. 0 g sample contained `Fe_(3)O_(4)` , Fe2O3 and inert materials . It was treated with an excess of aqueous Kl Solution in acidic medium,which reduced all the iron to `Fe^(+2)` ions. The resulting solution was diluted to 50.0 `cm^(3)` and a 10. 0 `cm^(3)` of it was taken. The liberated iodine in this solution required 7. 2 `cm^(3)` of 1. 0 M `Na_(2) S_(2)O_(3)` for reduction to iodide. The iodine from another 25. 0 `cm^(3)` sample was extracted , after which the `Fe^(+2)` ions was titrated against 1. 0 M `MnO_(4)^(-)` in acidic medium. The volume of `KMnO_(4)` solution used was found to be 4. 2 `cm^(3)` . Calculate the mass percentages of `Fe_(3)O_(4)` and of `Fe_(2)O_(3)` in the original mixture

Explanation

This problem can be done by two methods.
In the first method, we break up `Fe_(3) O_(4)` as an equimolar mixture of `FeO` and `Fe_(2) O_(3)`
Method ( 1)
`Fe_(3)O_(4) ` is` FeO. Fe_(2)O_(3)`
Equivalents of `Na_(2)S_(2) O_(3) = ( 1 xx 7.2 ) /( 1000) = 7.2 xx 10^(-3)`
Equivalents of `I_(2)` in 10cc `= 7.2 xx 10^(-3)`
Equivalents of `I_(2)` in 50 cc `= 7.2 xx 10^(-3) xx 5 = 3.6 xx 10^(-2)`
Since equivalents of `I_(2)` is equal to that of KI which in turn is equal to the total equivalents of `Fe_(2) O_(3) ( Fe_(2) O_(3)` in the free state and `Fe_(3) O_(3)` combined with `FeO)`
`:.` equivalents of `KMnO_(4)` solution
` = ( 4.2 xx 1 xx 5)/( 1000) = 2.1 xx 10^(-2)`
Since `KMnO_(4)` reacts with the total `Fe^(2+) ( Fe^(2+)` in FeO and `Fe^(2+)` that was produced by the action of Kl on `Fe_(2) O_(3))`
`:.` equivalents of total `Fe^(2-)` in 50ml
`= 2.1 xx 10^(-2) xx 2 = 4.2 xx 10^(-2)`
Since equivalent of `Fe^(2+)` produced from `Fe_(2)O_(3)` is equal to that of equivalents of `Fe_(2) O_(3)`
`:.` Equivalents of `FeO`
`= 4.2 xx 10^(-2) - 3.6 xx 10^(-2) = 6 xx 10^(-3)`
`:.` moles of `FeO = 6 xx 10^(-2)`
moles of `Fe_(2)O_(3)` combined with FeO `= 6 xx 10^(-3)`
total moles of `Fe_(2) O_(3) = ( 3.6 xx 10^(-2))/( 2)`
( because when `Fe_(2) O_(3) gt Fe^(2+) ` 'n' factor is 2 ) ` = 1.8 xx 10^(-2)`
moles of `Fe_(2) O_(3)` in the free state `= 1.8 xx 10^(_2) - 6 xx 10^(-3) = 1.2 xx 10^(-2)`
mass of `Fe_(3) O_(4) = 6 xx 10^(-3) xx 232 = 1.392g `
mass of `Fe_(3) O_(4) = 1.2 xx 10^(-2) xx 160 = 1.92 g `
percentage `Fe_(2) O_(4) = 17.4%`
percentage of `Fe_(2) O_(3) = 23.75 %`
Method 2`:`
Here we take `Fe_(3) O_(4)` as a single entity.
Equivalents of `Na_(2) S_(2) O_(3) =( 7.2 xx 1)/( 1000) = 7.2 xx 10^(-3) `
Equivalents of `I_(2) ` in 50 cc `= 7.2 xx 10^(-3) xx 5 = 3.6 xx 10^(_2)`
`:.` Equivalents of `Fe_(3) O_(4) +Fe_(2) O_(3) = 3.6 xx 10^(_2)`
Let us assume that the moles of `Fe_(3) O_(4)` in xg and that of `Fe_(2) O_(3)` is y g .
Since on reacting with KI both `Fe_(3) O_(4)` and `Fe_(2) O_(3)` give `Fe^(2+)` 'n' factor for both is two .
`:. 2x + 2y = 3.6 xx 10^(-3) ` ........(1)
Equivalent of `KMnO_(4) = ( 4.2 xx 1 xx 5 ) /( 1000) = 2.1 xx 10^(-2)` moles of `Fe^(2+)` in 50 mL `= 4.2 xx 10^(-2)`
Since the moles of `Fe_(3) O_(4)` are x, moles of `Fe^(2+)` produced from `Fe_(3) O_(4)` will be 3x and that produced from `Fe_(2) O_(3) `will be 2y
`:. 3x + 2y = 4.2 xx 10.2` .....(2)
`( 2) - ( 1) ` gives `x = 6 xx 10^(-3)`
`:. y = 1.2 xx 10^(-2)`
Solving this percentage of `Fe_(3) O_(4)` is 17.4 % and `Fe_(2) O_(3)` is 23.75%