Question

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.

Answer 1

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