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 was 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? M

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? M

First thing, I've never seen that type of question... but I'll try it ;)

Second thing, what is the purpose of the aluminium?

What is the reaction?

The equation for the reaction between Al(s) and Pb+2(aq) is:

2Al(s) + 3Pb+2(aq) 2 Al+3(aq) + 3Pb(s) .


I'm pretty sure that's it but I don't know its purpose

Ok the aluminium will displace the lead from the lead solution.

That means that in the solution of 100 ml, there were 0.016 mol of Pb^{2+} since all the aluminium reacted with the solution. From that, can you find the molarity of the Pb^{2+} in the 100 ml solution?

do I change the ml to liter and then divide 0.016 by that?

M = mole /over liter

so M = 0.016/0.1 but that equals 0?

No, 0.016/0.1= 0.16 mol

So, M = 0.16 mol/L or 0.16 mol/dm^-3