Answer:
1.02 g is the theoretical yield for PbCO₃
Explanation:
The reaction is
Pb(CH₃COO)₂ + CO₂ + H₂O → PbCO₃ + 2CH₃COOH
1 mol of lead(II) acetate react with 1 mol of carbon dioxide and 1 mol of water toproduce 1 mol of lead(II) carbonate and 2 moles of acetic acid.
With the two mass of the reactants, let's find put the limiting reactant.
1.25 g / 325.29 g/mol = 3.84×10⁻³ moles
5.95 g / 44 g/mol = 0.135 moles
Certainly, the limiting reactant is the lead(II) acetate. Ratio is 1:1, so if I have 0.135 moles of dioxide, I need the same amount of acetate. I only have 3.84×10⁻³ moles, that's why this is the limiting reactant.
Ratio is 1:1 too, with the lead(II) carbonate so 3.84×10⁻³ moles of acetate would produce 3.84×10⁻³ moles of carbonate.
Let's convert the moles to mass:
3.84×10⁻³ mol . 267.2 g/ mol = 1.02 g
