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 = ? %
(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?
3 answers
so you divide the molar mass by 2 and then multiply by fe's molar mass. why divide by 2?
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.
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.
2 answers