Question

Can someone help me please?

If 75.0 g of silderite ore (FeCO3) is heated with an excess of oxygen, 45.0 g of ferric oxide (Fe2O3) is produced.

4feCO3(s) + O2(g) -> 2fe2O3(s) + 4CO2(g)

what is the percent of yield of this reaction?,

Answer 1

The limiting reactant is FeCO3.
Molar mass of FeCO3 = 55.845 + 12 + 48 = 115.845
Number of moles of FeCO3 = 75/ 115.845 = 0.6475
But from the chemical reaction 4 moles of FeCO3 reacts with 2 moles of
Fe2O3. If we divide the number of molecules of our desired product, Fe2O3,
by the number of molecules of your limiting reactant FeCO3. We obtain our
theoretical yield. So we have 1/2 * 0.6475 = 0.3235 moles of Fe2O3
Molar mass of Fe2O3 = 159.7. Hence mass of Fe2O3 = 159.7 * 0.3235 =
51.66.
Percentage yield = Actual yield/ theoretical yield * 100
Percentage yield = 45/51.66 * 100 = 87%