Question

A mixture of FeO and FeO_{​​​​3} when heated in air to constant weight gains 5% in weight. Find out composition of mixture.

Plz explain the reaction involved in the process along with the equation.

Thanks in advance ...☺

Answer 1

Let us consider weight of FeO & Fe₃O₄ be 'a' & 'b' gm respectively.

Consider the following reactions.

2 FeO + 1/2 O₂ ------> Fe₂O₃

Fe₃O₄ + 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

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

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

Assuming total weight i.e. (a+b) = 100. -------- (3)

As there is an increase of 5 % we get, from eqn. 1,2 & 3 .

(160 x a /144) +160 x 3b/464 = 105

solving we get,

a= 21.06 & b = 78.94

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