I've been working at this problem in every way I can think of, but I am not coming up with a correct answer.
A correct equation I calculated for the problem: Pb(ClO3)2 + 2NaI = PbI2 + 2NaClO3
The part of the question I am having trouble solving is: What mass of precipitate will form if 1.50L of the concentrated Pb(ClO3)2 is mixed with 0.650L of 0.190 NaI? (Assuming the reaction goes into completion)