Potassium iodide and lead(II) nitrate solutions react together to form a precipitate of lead(II) iodide: 2KI(aq) + Pb(NO3)2(aq) → PbI2(s) + 2KNO3(aq) In each of the following cases, carry out the calculations to determine the quantities required. a If 1.0 mol of potassium iodide reacts with 1.0 mol of lead(II) nitrate, determine which reactant is in excess and by how many moles. b If 0.50 mol of potassium iodide reacts with 2.0 mol of lead(II) nitrate, determine which reactant is in excess and by how many moles. c If 1.00 g of lead(II) nitrate reacts with 1.50 g of potassium iodide, determine which reactant is in excess and the mass of lead(II) iodide that forms. d If 50.0 mL of 1.00 M lead(II) nitrate solution reacts with 75.0 mL of 0.500 M potassium iodide solution, determine which reactant is in excess the mass of lead(II) iodide that forms