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