A mixture containing only Al 2 O 3 (FM101.96) and Fe 2 O 3 (FM 159.69) weighs 2.

Question

A mixture containing only Al2O3 (FM101.96) and Fe2O3 (FM 159.69) weighs 2.019g.When heated under a stream of H2,Al2O3 is unchanged, butFe2O3 is converted to metallic Fe plusH2O(g). If the residue weighs 1.774g, what isthe weight % of Fe2O3 in the originalmixture? I have no idea what the residue consistsof, I tried just figuring the % by assuming the entire 1.774g wasFe, but the answer was wrong, so do I subtract the mass of thewater, or is there also Aluminum in the residue? I have no ideawhere to even start! BTW, the answer is supposed to be 40.4%wt. A mixture containing only Al2O3 (FM101.96) and Fe2O3 (FM 159.69) weighs 2.019g.When heated under a stream of H2,Al2O3 is unchanged, butFe2O3 is converted to metallic Fe plusH2O(g). If the residue weighs 1.774g, what isthe weight % of Fe2O3 in the originalmixture? I have no idea what the residue consistsof, I tried just figuring the % by assuming the entire 1.774g wasFe, but the answer was wrong, so do I subtract the mass of thewater, or is there also Aluminum in the residue? I have no ideawhere to even start! BTW, the answer is supposed to be 40.4%wt.

Explanation / Answer

residue is Al2O3 and Fe All the Oxygen from Fe2O3 is reacted (hence Fe is left over) (2.019 - 1.774) g O *(1 mol O/16 g) *(1 mol Fe2O3/3 mol O)*(159.69g/1mol Fe2O3) = 0.815 grams Fe2O3 0.815 grams Fe2O3 /2.019 total grams = 0.404 Thus 40.4 %

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