If you immerse aluminum metal in a solution of copper(II) sulfate, all you will get is wet aluminum. There will be no reaction. That's because aluminum metal is always covered in a thin layer of aluminum oxide, Al2O3, that prevents any reaction. This is referred to as "passivation" and the Al2O3 is called a "passivating layer."
Of course, according to the activity series, aluminum metal should "replace" copper ions in solution, but that doesn't take into account the reality of the passivating layer on aluminum. To make the reaction occur, all that's needed is a little hydrochloric acid (or some other source of chloride ion) which will result in the breakdown of the Al2O3 layer and produce AlCl4^- ions and expose the aluminum metal which will react with Cu2+ ions. What would be even easier would be to use CuCl2 instead of CuSO4.
Use the unit-factor method. If you can compute molar masses correctly, the unit-factor method is practically "idiot-proof."
2Al(s) + 3CuSO4(aq) --> Al2(SO4)3(aq) + 3Cu(s)
23.0g........ ?g
23.0g Al x (1 mol Al / 27.0g Al) x (3 mol CuSO4 / 2 mol Al) x (159.6g CuSO4 / 1 mol CuSO4) = 204g CuSO4
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Al + CuSO4.....
If you immerse aluminum metal in a solution of copper(II) sulfate, all you will get is wet aluminum. There will be no reaction. That's because aluminum metal is always covered in a thin layer of aluminum oxide, Al2O3, that prevents any reaction. This is referred to as "passivation" and the Al2O3 is called a "passivating layer."
Of course, according to the activity series, aluminum metal should "replace" copper ions in solution, but that doesn't take into account the reality of the passivating layer on aluminum. To make the reaction occur, all that's needed is a little hydrochloric acid (or some other source of chloride ion) which will result in the breakdown of the Al2O3 layer and produce AlCl4^- ions and expose the aluminum metal which will react with Cu2+ ions. What would be even easier would be to use CuCl2 instead of CuSO4.
Use the unit-factor method. If you can compute molar masses correctly, the unit-factor method is practically "idiot-proof."
2Al(s) + 3CuSO4(aq) --> Al2(SO4)3(aq) + 3Cu(s)
23.0g........ ?g
23.0g Al x (1 mol Al / 27.0g Al) x (3 mol CuSO4 / 2 mol Al) x (159.6g CuSO4 / 1 mol CuSO4) = 204g CuSO4
Find molar mass of CuSO4 = A .. Find molar mass of Al (just its atomic mass) = B
23 / [2A] X [3B] <<< use a calculator
Molar mass of Al = 27.0 g/mol
Molar mass of CuSO₄ = (63.5 + 32.1 + 16.0×4) g/mol = 159.6 g/mol
2 Al + 3 CuSO₄ → Al₂(SO₄)₄ + 3 Cu
Mole ratio Al : CuSO₄ = 2 : 3
No. of moles of Al = (23 g) / (27.0 g/mol) = 23/27 mol
No. of moles of CuSO₄ needed = (23/27 mol) × (3/2) = 23/18 mol
Mass of CuSO₄ needed = (23/18 mol) × (159.6 g/mol) = 204 g