Given the following reaction:

Question

4Fe + 3O₂ → 2Fe₂O₃

How many grams of Fe₂O₃ will be produced from 24.0g of Fe and 5.00g of O_2? Round to the three significant figures.

Answer 1

HI Kala!

You need to calculate the limiting reactant and use the lesser number of grams produced of Fe2O3 from each reactant:

1) From 24g Fe:

24 g Fe x (1 mol Fe/55.85 g Fe) x (2 mol Fe₂O₃/4 mol Fe) x (159.7 g Fe₂O₃/1 mol Fe₂O₃) = 34.3 g Fe₂O₃

2) From 5g O₂:

5 g O₂ x (1 mol O₂/32g O₂) x (2 mol Fe₂O₃/3 mol O₂) x (159.7 g Fe₂O₃/1 mol Fe₂O₃) = 16.6 g Fe₂O₃

Since 16.6g is lower we use that value as oxygen is the limiting reactant and Fe is in excess.