Mass of Al wire before reaction = 3.96 g Mass of Al wire after reaction = 3.65 g Mass of Al lost

Mass of Al wire before reaction = 3.96 g

Mass of Al wire after reaction = 3.65 g
Mass of Al lost = 0.31 g
Moles of Al lost = 0.011 moles
Mass of Pb + filter paper = 4.26 g
Mass of filter paper = 0.92 g
Mass of Pb = 3.34 g
Moles of Pb formed = 0.016 moles
Recall that there were 100 mL of solution. 0.016 moles of Pb were removed from the solution.
What was the [Pb+2] for the saturated solution of PbCl2?
Recall that there was 100 mL of solution. 0.016 moles of Pb were removed from the solution. Also recall the original equilibrium:
PbCl2(s) Pb+2(aq) + 2(Cl-)(aq)
Since there are 2 Cl ions formed for every Pb ion, what was the [Cl-] for the saturated solution of PbCl2?
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