Question

A Mixture Of FeO And Fe₂O₃ Were Heated In Air To A Constant Mass . It Was Found To Gain 10% In Its Mass . Calculate Its Percentage Composition Of The Original Mixture.

Answer 1

The following reactions take place :
2FeO + 1/2 O₂ → Fe₂O₃
2Fe₃O₄ + 1/2 O₂ → 3Fe₂O₃

Let the weight of FeO be X and weight of Fe₃O₄ be Y
X + Y = 100 ------------(1)
T)(hus,
2 x 72 g of FeO gives Fe₂O₃ = 160 g
X g of FeO gives Fe₂O₃ = [(160 x X) / 144] g
2 x 232 g of Fe₃O₄ gives Fe₂O₃ = 3 x 160 g
Y g of Fe₃O₄ gives Fe₂O₃ = [(3 x 160 x Y) / (2 x 232)] g
Therefore,
$\frac{160X}{144}+\frac{3\times 16\times Y}{464}=110------\left(2\right)$
Solving the equations (1) and (2),
We get,
X = 1.1 %
Y = 98.9 %