Question

"How many grams of the precipitate will form if 25.5 mL of 4.5 M solution of
lead (II) nitrate is allowed to react with 35.5 mL of 3.0 M solution of
potassium chloride? What is the concentration of the excess reactant after the reaction has reached completion?"

I got that the limiting is KCl and the excess is PbCl2. How do I find the concentration after the experiment? I get 4.5M, but I feel that's not right.

Answer 1

that's what you start with
... some of it is precipitated out

also, the volume of the solution is more than doubled

(remaining PbCl2) / solution volume

Answer 2

millimoles = mmols.
Pb(NO3)2 + 2KCl ==> PbCl2 + 2KNO3

mmols Pb(NO3)2 = mL x M = 25.5 x 4.5 = 114.75
mmols PbCl2 formed if you had all of the KCl you needed = 114.25

mmols KCl = mL x M = 35.5 x 3.0 = 106.5
mmols PbCl2 if you had all of the Pb(NO3)2 you needed = 53.25.

So KCl is the limiting reagent, you will form 53.25 mmols (0.05325 mols) PbCl2 or g PbCl2 = mols PbCl2 x molar mass PbCl2 = ? g PbCl2.

So you use all of the KCl. How much Pb(NO3)2 is used? That's 53.5 mmols KCl x [1 mol Pb(NO3)2/2 mols KCl)] = 106.5 x 1/2 = 53.25 mmols Pb(NO3)2 used. How much Pb(NO3)2 is left? That's 114.75-53.25 = 61.5 mmols Pb(NO3)2 not used. Then you have mmols/mL = 61.5/total mL = 61.5/61.0 mL = M Pb(NO3)2 in solution which is the excess reagent and that's what the problem asked, not (PbCl2).
Check all of these figures.