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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