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`        A MIXTURE OF FeO AND Fe3O4 WAS HEATED IN AIRR TO A CONSTANT MASS. IT WAS FOUND TO GAIN 10% IN ITS MASS.CALCULATE THE PERCENTAGE COMPOSITION OF THE ORIGINAL MIXTURE.`
7 years ago

Srinivas Rao
11 Points
```										85.5% of FeO and 14.5% of Fe3O4 on roasting shows 10% total increase in mass. Solution for the problem is as follows:

Fe3O4 can be regarded as FeO.Fe2O3.
Let x % of FeO and (100-x) % of Fe3O4 (Mol wt.232) be present in the mixture.
Upon roasting,
FeO (Mol wt.72) undergoes oxidation to half a mole of Fe2O3 (Mol wt.160) and FeO part of Fe3O4 also undergoes oxidation into half a mole of Fe2O3.
Thus,
72amu of FeO becomes 80amu of ½ Fe2O3 with 11.11% increase in mass
232amu of FeO.Fe2O3 becomes 240amu of (½ Fe2O3) Fe2O3 with an increase of 3.448% increase in mass.
Given, total increase in mass is 10%.
So, 11.11x + 3.448(1-x) =10
7.663x = 6.552
x = 0.855 or 85.5% of FeO
(1-x) = 14.5% of Fe3O4
```
7 years ago
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