Help plsss - Potassium iodide and lead(II) nitrate solutions react together to form solid lead(II) iodide and aqueous potassium nitrate: 2Kl(aq) + Pb(NO3)2(aq) → Pbl (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 mol.
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 mol.
c) If 1.00g of lead(II) nitrate reacts with 1.50g of potassium iodide, determine which reactant is in
excess and the mass of lead(II) iodide that forms.