Question

If 1176 grams of FeS₂ is allowed to react with 704 grams of O₂ according to the following unbalanced equation, how many grams of Fe₂O₃ are produced?

FeS₂ + O₂ → Fe₂O₃ + SO₂

Answer 1

Ans:

Mass of FeS₂ taken = 1176 g

Molar mass of FeS₂ = 119.98 g/mol

No. of moles of FeS₂ taken = 1176/119.98 = 9.80 mol

Mass of O₂ taken = 704 g

Molar mass of O₂ = 32 g/mol

No. of moles of O₂ taken = 704/32 = 22.00 mol

Balanced reaction is: 4 FeS₂ + 11 O₂ → 2 Fe₂O₃ + 8 SO₂

We can see from the above balanced reaction that for 4 mol of FeS₂ we need 11 mol of O₂.

So, for 9.80 mol of FeS₂, moles of O₂ needed = (9.80 x 11)/4 = 26.95 mol. But we have only 22 mol of O₂. Meaning, O₂ is a limiting reagent.

11 moles of O₂ reacts with 4 moles of FeS₂. So, 22 moles of O₂ reacts with = (22 x 4)/11 = 8 moles of FeS₂. Meaning, of the 9.80 moles of FeS₂ only 8 moles will react.

We can see from the above balanced reaction that 4 moles of FeS₂ produce 2 moles of Fe₂O₃. So, 8 moles of FeS₂ produce 4 moles of Fe₂O₃.

Molar mass of Fe₂O₃ = 159.69 g/mol

Mass of 4 moles of Fe₂O₃ = 159.69 g/mol x 4 mol = 638.76 g

Mass of Fe₂O₃ produced = 638.76 g