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(Iron ore is impure Fe_2O_3. When Fe_2O_3 is heated with an excess of carbon (coke), metallic iron and carbon monoxide gas are produced. From a sample of ore weighing 950kg , 533kg of pure iron is obtained). What is the mass percent by mass, in the ore sample, Fe_2O_3, assuming that none of the impurities contain Fe?

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  1. 533 kg Fe x (Fe2O3/2*Fe) =
    533 kg x (159.69/2*55.85) = ? kg Fe2O3 and substitute this into the below.

    %Fe2O3 = (?mass Fe2O3/950kg)*100 = ? %

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  2. so you divide the molar mass by 2 and then multiply by fe's molar mass. why divide by 2?

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  3. Because thee are two Fe atoms in one molecule of Fe2O3.
    533 kg Fe x (1 mol Fe/55.85 g Fe) x (1 mol Fe2O3/2 mols Fe) x (159.69 g Fe2O3/1 mol Fe2O3) = ?kg Fe2O3. Note all of the units cancel and you are left with
    533 x [159.69/(2*55.85)] = ? kg Fe2O3.

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