Question

Given the reaction: 4 al(s) + 3 o2(g) → 2 al2o3(s) what is the minimum number of grams of oxygen gas required to produce 1. 00 mole of aluminum oxide?.

Answer 1

The reaction requires 3 mol O₂ for every 2 mol Al₂O₃ that are produced. This means we need 1.50 mol O₂ to produce 1.00 mol Al₂O₃, which has a mass of about

(1.50 mol) (31.998 g/mol) ≈ 48.0 g