Mass of Al wire before reaction = 3.96
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
What was the [Pb2] 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
equilibrium:
PbCl2(s) — Pb 2(aq) + 2(CH)(aq)
Since there are 2 Clions formed for every Pb ion, what was the (CI) for the saturated solution of PbCl2