A 10.2 g mixture of Aluminum and Magnesium was treated with excess hydrochloric acid, releasing 11.2 liters of hydrogen gas at STP. Determine the mass of Aluminum in the original mixture.
- 5.4 g
- 4.8 g
- 2.7 g
- 7.5 g
📝 Batafsil tushuntirish
n(H₂) = 11.2 / 22.4 = 0.5 mol.Let x = mol Al, y = mol Mg.
1) 27x + 24y = 10.2
2) 1.5x + y = 0.5 (from 2Al+6HCl→2AlCl₃+3H₂ and Mg+2HCl→MgCl₂+H₂)
From (2): y = 0.5 - 1.5x.
Substitute into (1): 27x + 24(0.5 - 1.5x) = 10.2
27x + 12 - 36x = 10.2
-9x = -1.8 => x = 0.2 mol.
Mass Al = 0.2 * 27 = 5.4 g.