Solid aluminum and gaseous oxygen react in a combination reaction to produce aluminum oxide: 4Al (s) + 3O₂ (g) → 2Al₂O₃ (s)
The maximum amount of Al₂O₃ that can be produced from 2.5 g of Al and 2.5 g of O₂ is ________ g.

Question

Quizplus · Accepted Answer

1) First we need to figure out which reactant is the limiting reagent by using stoichiometry.
2.5 g Al * (1 mol Al/26.98 g Al) * (2 mol Al2O3/4 mol Al) * (101.96 g Al2O3/mol Al2O3) = 9.42 g Al2O3
2.5 g O2 * (1 mol O2/32 g O2) * (2 mol Al2O3/3 mol O2) * (101.96 g Al2O3/mol Al2O3) = 7.51 g Al2O3
Aluminum is the limiting reagent since it produces less moles of Al2O3 . 
2) We can now use the amount of Al to calculate how much Al2O3 is produced.
2.5 g Al * (1 mol Al/26.98 g Al) * (2 mol Al2O3/4 mol Al) * (101.96 g Al2O3/mol Al2O3) = 9.42 g Al2O3
Therefore, the maximum amount of Al2O3 that can be produced is 9.42 g.

Solid Aluminum and Gaseous Oxygen React in a Combination Reaction