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"
