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.

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