Answer:
The correct answer is 15.80 grams.
Explanation:
The reaction taking place in the given question,
Pb(CH₃COO)₂ + Na₂SO₄ ⇒ PbSO₄ + 2NaCH₃COO
The number of moles can be calculated by using the formula,
n = weight / molecular mass
Based on the given question, the weight of lead (II) acetate is 23.81 grams and the weight of sodium sulfate is 7.410 grams.
The number of moles of Pb(CH₃COO)₂ is,
n = 23.81 g / 325.29 g/mol = 0.0732 moles
The number of moles of Na₂SO₄ is,
n = 7.410 g / 142.04 g/mol = 0.0521 moles
As one mole of lead (II) acetate needs one mole of sodium sulfate. Therefore, 0.0732 moles of lead (II) acetate needs 0.0732 moles of sodium sulfate.
However, as sodium sulfate is less, that is, 0.0521, therefore, Na₂SO₄ is a limiting reactant.
One mole of sodium sulfate produces one mole of PbSO₄. So, 0.0521 moles of Na₂SO₄ produces 0.0521 moles of PbSO₄.
Now the mass of PbSO₄ is,
mass = moles × molecular mass
mass = 0.0521 × 303.26 g/mol
mass = 15.80 grams.
