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 since all the aluminium reacted with the solution. From that, can you find the molarity of the 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