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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.

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.

Grade:11

1 Answers

Srinivas Rao
11 Points
11 years ago

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

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