Question

A mixture of FeO and Fe2O3 when heated in air to a constant weight gains8% in its weight. The % of Fe2O3 in the given mixture is _________
(Atomic mass of Fe= 56 amu)

Answer 1

Explanation:

Let, weight of FeO = x; Weight of Fe3O4 = y ∴ x + y = 100 …(1) ∴ 2 × 72 gm of FeO give Fe2O3 = 160 gm ∴ x gm FeO gives Fe2O3 = 160 x / 44 gm 2 × 232 gm of Fe3O4 gives Fe2O3 = 3 × 160 gm ∴ y gm Fe3O4 gives Fe2O3 = 3 x 160y /2 x 232 gm ∴ (160 x / 44) + (3 x 160y /2 x 232) = 105 …(2) Solving equation (1) & (2) x = 20.25 gm; y = 79.95 gm

Answer 2

Answer:

20.25gm and 79.95gm

Explanation: