In this lab we saw the reaction of lead (II) nitrate with potassium iodide to form a lead (II) iodide precipitate and aqueous potassium nitrate. Pb(NO3)2 + 2KI ----> PbI2 + 2KNO3(aq) There was 3.31 grams Lead Nitrate, 1.66 grams Potassium Iodide, and 0.48 grams lead (II) iodide precipitate. What I need to know is the following (please show your work so I can learn how to do it.): Question #1: HOw many moles of PbI2 should be produced if the solutions are mixed? Question #2: What mass of PbI2 should be produced? Question #3: What is the theoretical yield of the precipitate and the percent yield of lead (II) iodide formed. Please help me!!!!