Question

A mixture of FeO and Fe3O4 when heated in air to constant weight gains 5% in its weight. Find the composition of its initial mixture

Answer 1

Let weight of FeO & Fe₃O₄ be 'X' & 'y' gm respectively.
The reactions occurring a re as follows:

2 FeO + 1/2 O₂ ------> Fe₂O₃

2Fe₃O₄ + 1/2 O₂ ------> 3Fe₂O₃

Molar mass of FeO =144 g and that of Fe₂O₃ is 160 gms.
Thus 144 g FeO gives 160 g Fe₂O₃

'X' g FeO will give 160 x X/144 gm Fe₂O₃ --------------(1)

Similarly, weight of Fe₂O₃ formed by Y gm Fe₃O₄ = 160 x 3Y/464 --------(2)

Assuming total weight i.e. (X+Y) = 100. -------- (3)
As there is an increase of 5 % we get, from eqn. 1,2 & 3 .

(160 x X /144) +160 x 3y/464 = 105

solving we get,
X= 21.06 & Y = 78.94

Thus % of FeO = 21.06 % & that of Fe₃O₄ = 78.94 %