Question

When a solution of Lead Nitrate, Pb(NO₃)₂ , is mixed with a solution of sodium iodide, NaI, a precipitate of lead iodide, PbI₂ , is formed.

a) Write the chemical equation for this reaction showing the state of all reactants and products.

b) If 1 mol of Pb(NO₃)₂ reacts with 2 mols of NaI, what is the amount of PbI₂ that will be formed?

c) If 1 mol of Pb(NO₃)₂ reacts with 2 mols of NaI, which compound would be the limiting reagent?

d) For the mixture described in c), what is the amount of PbI₂ that forms?

e) Solution A contains 50.00 ml of 1.000 x 10^-1 mol x L^-1 Pb(NO₃)₂   and solution B contains 50.00 ml of 1.000 x 10^-1 mol x L^-1   NaI. Calculate the mass of Pb(NO₃)₂ in solution A and the mass of NaI in solution B.

f) When solution A and B described above are mixed, solid PbI₂ forms. Calculate the theoretical yield of PbI₂.

g) 1.021 g of solid PbI₂ was actually collected from the mixture above. Calculate the percent yield of PbI₂.

Answer 1