Aluminum oxide (used as an adsorbent or a catalyst for organic reactions) forms when aluminum reacts with oxygen. 4Al(s) + 3O₂(g) → 2Al₂O₃(s)
A mixture of 82.49 g of aluminum (ℳ= 26.98 g/mol) and 117.65 g of oxygen (ℳ = 32.00 g/mol) is allowed to react. What mass of aluminum oxide (ℳ = 101.96 g/mol) can be formed?

Question

Quizplus · Accepted Answer

First, calculate the moles of aluminum and oxygen. For aluminum: 82.49 g / 26.98 g/mol = 3.06 mol. For oxygen: 117.65 g / 32.00 g/mol = 3.68 mol. The reaction ratio is 4Al : 3O2 = 2Al2O3. Aluminum is the limiting reactant (3.06 mol Al would need 2.295 mol O2, but there's more O2 available). So, all aluminum reacts. The moles of Al2O3 formed = (3.06 mol Al) / (2 mol Al per mol Al2O3) = 1.53 mol Al2O3. Mass of Al2O3 = 1.53 mol * 101.96 g/mol = 155.8 g.

Aluminum Oxide (Used as an Adsorbent or a Catalyst for Organic