How much iron can be obtained from 50 kg of pyrite (FeS2)

To solve, we write down the process:

4FeS2 + 11O2 = 2Fe2O3 + 8SO2 – pyrite roasting, iron oxide was obtained (3);
Fe2O3 + 3CO = 2Fe + 3CO2 – OBP, iron is released.
Calculations:
M (FeS2) = 119.8 g / mol;

M (Fe) = 55.8 g / mol.

4. Determine the amount of the original substance:

Y (FeS2) = m / M = 50,000 / 119.8 = 417.4 mol.

5. Proportions:

417.4 mol (FeS2) – X mol (Fe2O3);

-4 mol -2 mol hence, X mol (Fe2O3) = 417.4 * 2/4 = 208.7 mol;

208.7 mol (Fe2O3) – X mol (Fe);

-1 mol -2 hence, X mol (Fe) = 208.7 * 2/1 = 417.4 mol.

Find the mass of the metal:
m (Fe) = Y * M = 417.4 * 55.8 = 23290 g = 23 kg 290 g

Answer: received iron weighing 23 kg 290 g