The balanced equation for the reaction of aqueous Pb(ClO3)2 with aqueous NaI is shown below.
Pb(ClO3)2 (aq) + 2NaI (aq) -> PbI2 (s) + 2NaClO3 (aq)
What mass of precipitate will form if 1.50 L of excess Pb(ClO3)2 is mixed with 0.100 L of 0.230 M NaI? Assume 100% yield and neglect the slight solubility of PbI2.