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3 January, 06:13

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.

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  1. 3 January, 07:47
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    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
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