If a solution containing 23.81 g of lead(II) acetate is allowed to react completely with a solution containing 7.410 g of sodium sulfate, how many grams of solid precipitate will be formed g

Question

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Respuesta :

Larus Larus · Answer 1 · 2020-06-29T04:17:33+02:00

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.