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A 43.32mL solution of 0.1456M Pb(NO3)2 solution is added to 44.34mL 0.289M KI solution. What will...

A 43.32mL solution of 0.1456M Pb(NO3)2 solution is added to 44.34mL 0.289M KI solution. What will be the mass of lead (II) iodide formed and what is the final concentration of the Pb2+ ion or Iion in solution whatever ion is in excess?

Pb(NO3)2(aq) + KI(aq) ---> PbI2(s) + KNO3(aq)

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