Iron oxide has formula `Fe_"0.94"O.` What fraction of Fe exists as `Fe^(+3)` ?

Explanation

Let there are 100 oxide ions and 94 iron ions.
`Fe^(+x):O^(-2)` Total negative charge on 100 oxide ions =-200
=94 :100
It means as per the concept of charge neutrality , 94 iron ions should have total charge of +200
let number of `Fe^(+3)`=x
`therefore` Number of `Fe^(+2)` =(94-x)
Total positive charge =3x-2(94-x)=200
On solving , x =12
%`Fe^(+3) =12/94xx100` =12.8%