Suppose an aqueous solution containing 1.25 g of lead (II) acetate is treated with 5.95 g of carbon diox - ide. Calculate the theoretical yield of lead carbonate.

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