How do I solve this problem,Please be detailed

Question

 A solution containing 5.00 g of lead(ll) acetate is mixed with a solution containing 3.00 g of ammonium iodide.

Pb(C2H3O2)2 (aq) + 2NH4I (aq) = PbI2 (s) + 2NH4C2H3O2 (aq)

a)How much lead(II) iodide is generated??

b) assuming the reaction occurred in a 1.0 L solution, what is the final concentration of the excess reactant ? 

Answer 1

a) First, we have to find the limiting reagent

5.00 g Pb(C2H3O2)2 • 1 mol Pb(C2H3O2)2/ 325.29 g Pb(C2H3O2)2 = 0.0153 mol Pb(C2H3O2)2

3.00 g NH4I • 1 mol NH4I/ 144.94 g NH4I = 0.0207 mol NH4I

If we assume that lead acetate is limiting, we would get the following equation based on the coefficients of the balanced reaction:

0.0153 mol Pb(C2H3O2)2 • 2 mol NH4I/ 1 mol Pb(C2H3O2)2 = 0.0306 mol NH4I

Since this number is bigger than the amount of ammonium iodide we start with, it means that ammonium iodide is the limiting reagent.

0.0207 mol NH4I • 1 mol PbI₂/2 mol NH₄I • 461.01 g PbI₂/1 mol PbI₂ = 4.77 g PbI₂

b) To find out how much excess reactant (in this case lead acetate) is left over, we need to find out how much we consume.

0.0207 mol NH₄I • 1 mol PbI₂/2 mol NH₄I = 0.0104 mol

0.0153 PbI₂ initial - 0.0104 PbI₂ consumed =0.005 mol PbI₂ remaining

since our volume is 1L, then concentration = 0.005/1 = 0.005 M PbI₂