Find the mass of alcl3 that is produced when 10. 0 grams of al2o3 react with 10. 0 g of hcl according to the following equation. Al2o3(s) + 6hcl(aq) → 2alcl3(aq) + 3h2o(aq)

The balanced reaction

Al₂O₃ (s) + 6 HCl (aq)   →   2 AlCl₃ (aq) + 3 H₂O (aq)

says that 1 mol Al₂O₃ reacts with 6 mol HCl to produce 2 mol AlCl₃.

We start with

• (10.0 g Al₂O₃) (1/101.96 mol/g) ≈ 0.0981 mol Al₂O₃

• (10.0 g HCl) (1/36.46 mol/g) ≈ 0.274 mol HCl

We can use up to

(0.274 mol HCl) (1 mol Al₂O₃ / 6 mol HCl) ≈ 0.0457 mol Al₂O₃

and consume all the HCl in the process (while also leaving an excess of about 0.0524 mol Al₂O₃) to produce approximately

2 (0.0457 mol Al₂O₃) (2 mol AlCl₃ / 1 mol Al₂O₃) ≈ 0.183 mol AlCl₃

This has a mass of about

(0.183 mol AlCl₃) (133.3 g/mol) ≈ 24.4 g AlCl₃