Chemistry: Stoichiometry

Question

Approximately (3.230x10^0) g of lead(II) acetate, Pb(CH₃COO)_2, is weighed out and placed in a (2.85x10^2) mL solution of 0.150-mol/L sodium hydroxide. Calculate the mass of precipitate produced in this reaction. 

Note: Your answer is assumed to be reduced to the highest power possible.

Answer 1

Let's look at the reaction taking place to produce the precipitate:

Pb(CH₃COO)₂(aq) + 2NaOH(aq) ==> Pb(OH)₂(s) + 2NaCH₃COO(aq)

Next, let's look at the amounts of each reactant that is present:

3.230x10⁰ g Pb(CH₃COO)₂ = 3.230 g x 1mol / 325 g = 0.0099 moles Pb(CH₃COO)₂

2.85x102 ml NaOH = 285 ml = 0.285 L x 0.150 mol/L = 0.0428 moles NaOH

Based on these values, Pb(CH₃COO)₂ is the limiting reactant and we can use that to find the theoretical yield of Pb(OH)₂:

0.0099 mols Pb(CH₃COO)₂ x 2 mols Pb(OH)₂ / mol Pb(CH₃COO)₂ = 0.0198 mols Pb(OH)₂

Converting this to grams, we have 0.0198 mols x 241 g/mol = 4.48 g Pb(OH)₂

This would be the mass of ppt formed, IF IT ALL WERE TO PRECIPITATE.

Since no data are provided concerning solubility and/or Ksp for Pb(OH)2, we will assume that all is precipitated.

I'm not entirely sure what is meant by "Your answer is assumed to be reduced to the highest power possible."