What mass of aluminum oxide is produced?

Question

In a reaction, 42.9 g of manganese(III) oxide reacts with 14.7 g of aluminum to produce manganese and aluminum oxide. If 29.9 g of manganese is produced, what mass of aluminum oxide is produced?

Answer 1

This is a Stoichiometry problem, pure and simple. Take either of the starting masses you are given and convert it to moles of that substance. Then convert to moles of Aluminum Oxide (Al₂O₃). In this case, it doesn't matter which reactant you choose as they both give you 0.272 moles of Aluminum Oxide (rounded to 3 sig figs).

The balanced equation for this problem is as follows:

Mn₂O₃ + 2 Al = 2 Mn + Al₂O₃

42.9 g Mn₂O₃ * mol Mn₂O₃ * mol Al₂O₃ * 101.96 g Al₂O₃ = 27.7 g Al₂O₃ produced

157.88 g Mn₂O₃ mol Mn₂O₃ mol Al₂O₃