#### a mixture of feo and fe3o4 when heated in air to constant ...

Molar mass of FeO =144 g and that of Fe 2 O 3 is 160 gms. Thus 144 g FeO gives 160 g Fe 2 O 3 'a' g FeO will give 160 x a/144 gm Fe 2 O 3(1) Similarly, weight of Fe 2 O 3 formed by 'b'gm Fe 3 O 4 = 160 x 3b/464 (2) Assuming total weight (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= b = Thus % of FeO = .