Consider the balanced equation of Ki (molar mass=166.0g/mol) reacting with Pb(NO3)2 to form a precipitate. 2 KI (aq) + Pb(NO3)2(aq) ---> PbI2(s)+2KNO3(aq)
What mass of PbI2 (molar mass = 461.0 g/mol) can be formed by adding 3.67g of KI to a solution of excess Pb(NO3)2?
