Consider the following reaction:
4FeS2(s) + 11O2(g) → 2Fe2O3(s) + 8SO2(g)
How many grams of Fe2O3(s) are produced from 500.0 g of FeS2(s) and 400.0 g of oxygen gas?
500g FeS2 = 4.17 moles
400g O2 = 12.5 moles
Since 4 moles of FeS2 needs 11 moles of O2, 4.17 moles will need 11.47 moles of O2
So, the FeS2 will be the limiting reactant, producing 2.08 moles of Fe2O3
how many grams is that?