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A mixture of FeO and FeO3 when heated in air to constant weight gains 5% in weight. Find out composition of mixture.

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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 

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