The composition of a simple of Wustrie is `Fe_(0.03)O`. What is the percentage of iron present as `Fe^(3+)` in total iron?

Explanation

In the given oxide 96 M-ions are associated with 100 - `O^(2-)` ions. Out of 96 M ions, suppose `M^(2+)` ion = x, then `M^(3+)` ion = 96 - x.
Total the charge on x `M^(2+)` ion = 2x
Total the c harge on (96 - x) `M^(3+)` ion
=3 (96 - x)
`therefore` Total the charge on M-ions
`=2x+3(96-x)`
Total the charge = Total -ve charge
`2x+3(96-x)=200`
`2x+288-3x=200`
x = 88
`therefore` % of `M^(2+)` ion `=88/(96)xx100=91.7%`
% of `M^(3+)` ion `=100-91.7=8.3%`