A 20 g mixture of Al and Al₂O₃ was treated with an excess of NaOH solution, releasing 6.72 L of gas at STP. What is the mass fraction of Al₂O₃ in the original mixture?
- 27%
- 54%
- 73%
- 46%
📝 Detailed explanation
Only Al reacts with NaOH to release H₂: 2Al + 2NaOH + 6H₂O → 2Na[Al(OH)₄] + 3H₂.Moles H₂ = 6.72 / 22.4 = 0.3 mol.
Moles Al = 0.3 × (2/3) = 0.2 mol.
Mass Al = 0.2 × 27 = 5.4 g.
Mass Al₂O₃ = 20 - 5.4 = 14.6 g.
Mass fraction = 14.6 / 20 = 0.73 or 73%.