For the reactions shown below, we added 2.25 mL of 0.0500 M HCl to a test tube containing one of the two cations (Ag+ or Pb2+) and recovered 0.0161 g of precipitate.
AgNO3(aq) + HCl(aq) → AgCl(s) + HNO3(aq)
Pb(NO3)2(aq) + 2 HCl(aq) → PbCl2(s) + 2 HNO3(aq)
How much precipitate in moles would be recovered theoretically if the ion was Pb2+?

Question

For the reactions shown below, we added 2.25 mL of 0.0500 M HCl to a test tube containing one of the two cations (Ag+ or Pb2+) and recovered 0.0161 g of precipitate.
AgNO3(aq) + HCl(aq) → AgCl(s) + HNO3(aq)
Pb(NO3)2(aq) + 2 HCl(aq) → PbCl2(s) + 2 HNO3(aq)
How much precipitate in moles would be recovered theoretically if the ion was Pb2+?

cometailcane · Answer

So I found how many moles of Ag+ which was 0.0001125. 
For Pb I did 2.25 mL x 0.0500 M HCl = 0.1125 mmol = 0.0001125 mol. And since the ratio is 1 to 2 I divided by 2 and got 0.05625 but that is not the correct answer

cuanchi · Answer

0.05625 incorrect
0.0001125/2 = 5.625 x 10^-5

cometailcane · Answer

What about this? How much precipitate in grams would be recovered theoretically if the ion was Ag+?

cometailcane · Answer

I thought all I had to do was 0.0001125 x the molar mass of Ag

cuanchi · Answer

x the molar mass of AgCl