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Oxidation no of O = -2
Oxidation no of Fe were = +2
composition would have been FeOFe is in the ratio Fe0.93O1.00
Thus , Fe has some atoms in +2 oxidation state and some in +3
Let the amount of Fe(II) be x and Fe(III) be ythenx +y =0.93 ...........( given composition)2x + 3y = 2 ..........(Charge of Fe (2+) and charge of Fe (+3) equal to charge of the compound }solving them we get x = 0.79 and y = 0.14now,% of Fe(III) = 0.14/0.93 * 100 = 15% :)15%
Regards
Arun (askIItians forum expert)
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