I don't know if i am solving for the number moles of NaI correctly, i multiply .21 by 1.67, is that incorrect? i know that to solve this you have to find the limiting reagent, but i think the problem might be that i am solving for moles wrong.

Question

If 210 mL of 1.67 M aqueous NaI and 156 g of Pb(NO3)2 are reacted stoichiometrically according to the balanced equation, how many moles of solid PbI2 are produced?

Pb(NO3)2(aq) + 2NaI(aq) → PbI2(s) + 2NaNO3(aq

Answer 1

moles NaI = M x L.
moles Pb(NO3)2 = grams/molar mass.