200cm3 of each of 0.1m of solutions of lead (ii) trioxonitrate (v) and hydrochloric acid were mixed. Assuming that lead (ii) chloride is completely insoluble, calculate the mass of lead (ii) chloride that will be precipitated.

Answer 1

Pb(NO3)2 + 2HCl ==> PbCl2 + 2HNO3
Assuming you meant 0.1M and not 0.1m,
millimols Pb(NO3)2 = mL x M = ?
millimols HCl = mL x M = ?
Determine the limiting reagent and you should see that to be HCl so that means you heave 10 millimols Pb(NO3)2 excess. Convert to mols, then grams PbCl2 = mols x molar mass PbCl2 = ?

Ther is an excess of Pb^2+ and that is a common ion. You can see how that affects the solubility of PbCl2 but I doubt you need to go through that calculations. If you want to go through it, it is done this way.
............PbCl2 ==> Pb^2+ + 2Cl^-
I...........solid......0.......0
C...........solid......x.......2x
E...........solid......x.......2x

Ksp = (Pb^2+)(Cl^-)^2
Look up Ksp for PbCl2.
(Pb^2+) = 10 millimols/400 mL = ?. Put that in for Pb^2+ in the Ksp expression above, solve for Cl^-, take half of that to give solubility of PbCl2, convert to grams PbCl2, subtract from the solid you calculated above. As I stated above, I doubt this will be significant. The difference is the amount in grams of PbCl2 that is produced.