VERY LONG ANSWER
Here's what you're working with. Sodium iodide will fully dissociate in aqueous solution to give
#NaI_((aq)) rightleftharpoons Na_((aq))^(+) + I_((aq))^(-)#
This means that the concentration of #I^(-)# ions will be equal to the concentration of the sodium iodide in the final solution. The moles of #NaI# added are
#n = C * V = 220 * 10^(-3)"L" * "0.0255 M" = 5.61 * 10^(-3)"moles"#
The concentration of sodium iodide in the final solution will be
#C_("iodide") = n/V_("final") = (5.61 * 10^(-3)"moles")/((220 + 340) * 10^(-3)"L") = "0.01 M"#
Therefore, the concentration of #I^(-)# ions will be: #[I^(-)] = "0.01 M"#.
Lead (II) chloride however will not fully dissociate in aqueous solution.The solution is unsaturated because the minimum concentration of #Pb^(2+)# ions in solution for a precipitate to form is #"0.016 M"#. Since your starting concentration of #PbCl_2# is less than that, the lead (II) chloride will dissolve in solution and not remain solid.
As a result, the concentration of the #Pb^(2+)# ions will be equal to that of the lead (II) chloride, or #[Pb^(2+)] = "0.005 M"#
SIDE NOTE. You can try and determine that through calculation by using the ICE table method (more here: http://en.wikipedia.org/wiki/RICE_chart) and #PbCl_2#'s solubility product constant, #K_(sp) = 1.6 * 10^(-5)#.
For the final solution, the concentration of #Pb^(2+)# ions will be equal to
#n_("lead (II)") = 340 * 10^(-3)"L" * "0.005 M" = 1.7 * 10^(-3)"moles"#
#C_("lead (II)") = n/V_("final") = (1.7 * 10^(-3)"moles")/(560 * 10^(-3)"L") = "0.003 M"#
This will be the final concentration of #Pb^(2+)# ions in solution: #[Pb^(2+)] = "0.003 M"#.
The reaction that is of interest for you will be
#2I_((aq))^(-) + Pb_((aq))^(2+) rightleftharpoons PbI_(2(s))#
However, a precipitate will only form if the concentrations of the two ions are large enough. So,
#Q_c = [Pb^(2+)] * [I^(-)]^(2) = "0.003 * 0.01^(2) = 3.0 * 10^(-7)#
Since this time #Q_c > K_(sp) = 8.3 * 10^(-9)#, #PbI_2# will form as a precipitate in the final solution.
Now you have to find the limiting reagent in order to determine how much salt is produced. Since the number of #I^(-)# moles is equal to #5.61 * 10^(-3)#, and the number of #Pb^(2+)# moles is equal to #1.7 * 10^(-3)#, look at the mole ratio you have between the two ions.
One mole of #Pb^(2+)# ions needs 2 moles of #I^(-)# ions #-># you have a #1:2# mole ratio between the two. This means that #Pb^(2+)# will be the limiting reagent, since
#2 * 1.7 * 10^(-3) = 3.4 * 10^(-3) < 5.61 * 10^(-3)#
This means that the number of moles of #PbI_2# produced will be
#1.7 * 10^(-3)"moles" * ("1 mole Pb"^(2+))/("1 mole PbI"_2) = 1.7 * 10^(-3)"moles"#
Therefore,
#1.7 * 10^(-3)"moles" * ("461 g")/("1 mole") = "0.78 g"# #PbI_2#
Now for the ions that remain in solution. Since #Pb^(2+)# was the limiting reagent, it was fully consumed in the reaction; this means that its remaining concentration in solution will be zero. The concentration of the #I^(-)# ions is determined by the excess moles
#n_("excess"I^(-)) = 5.6 * 10^(-3) - 2 * 1.7 * 10^(-3) = 2.2 * 10^(-3)"moles"#
Thus, the remaining concentration of #I^(-)# ions will be
#[I^(-)] = (2.2 * 10^(-3)"moles")/(560 * 10^(-3)"L") = "0.0040 M"#
