(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?
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.
To find the mass percent of Fe2O3 in the ore sample, we need to determine the amount of Fe2O3 present in the 533 kg of pure iron obtained from the sample.
First, let's determine the mass of Fe2O3 in the pure iron obtained. Since Fe2O3 is the only source of iron in this reaction, the mass of pure iron obtained is equal to the mass of Fe2O3.
Therefore, the mass of Fe2O3 in the pure iron obtained is 533 kg.
Now, we can calculate the mass percent of Fe2O3 in the ore sample using the formula:
Mass percent = (mass of Fe2O3 / mass of ore sample) x 100
Given:
Mass of Fe2O3 in the pure iron = 533 kg
Mass of ore sample = 950 kg
Substituting these values into the formula:
Mass percent = (533 kg / 950 kg) x 100
Simplifying:
Mass percent = 0.561 x 100
Mass percent = 56.1%
Therefore, the mass percent by mass of Fe2O3 in the ore sample is 56.1%.
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 = ? %