single displacement

Question

If a lead (III) sulfate solution is in a 0.3 M solution, what volume would you have to add to unlimited aluminum to produce 5 g of pure lead in a single-displacement reaction?

Answer 1

Here is the balanced equation for the reaction in question.

2Al(s) + 3PbSO₄ ==> Al₂(SO₄)₃ + 3Pb(s)

NOTE: 3PbSO₄ is NOT soluble in water so this reaction will not proceed readily. If you want to assume it is soluble and that the reaction will proceed (an invalid assumption), then proceed as follows:

5 g Pb x 1 mol Pb/207 g = 0.0242 moles

mols PbSO₄ needed = 0.0242 moles Colton 0.3 M PbSO₄ needed = 0.024 mol x 1L/0.3 mol = 0.081 L = 81 mls = 80 mls (1 sig fig)