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