Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction
Pb(NO3)2(aq)+2NH4I(aq)--------PbI2(s)+2NH4NO3(aq)
What volume of a 0.170 M NH4I solution is required to react with 963 mL of a 0.440 M Pb(NO3)2 solution?
How many moles of PbI2 are formed from this reaction?