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The mass of the precipitate, BaSO₄ obtained from the given reaction is 18.64 grams

How to determine the mole Na₂SO₄

  • Volume of Na₂SO₄ = 300 mL = 300 / 1000 = 0.3 L
  • Molarity of Na₂SO₄ = 0.3 M
  • Mole of Na₂SO₄ =?

Mole = Molarity x Volume

Mole of Na₂SO₄ = 0.3 × 0.3

Mole of Na₂SO₄ = 0.09 mole

How to determine the mole of BaCl₂

  • Volume of BaCl₂ = 200 mL = 200 / 1000 = 0.2 L
  • Molarity of BaCl₂ = 0.4 M
  • Mole of BaCl₂ =?

Mole = Molarity x Volume

Mole of BaCl₂ = 0.4 × 0.2

Mole of BaCl₂ = 0.08 mole

How to determine the limiting reactant

Balanced equation

Na₂SO₄(aq) + BaCl₂(aq) —> BaSO₄(s) + 2NaCl(aq)

From the balanced equation above,

1 mole of BaCl₂ reacted with 1 mole of Na₂SO₄.

Therefore,

0.08 mole of BaCl₂ will also react with 0.08 mole of Na₂SO₄.

From the above illustration, we can see that only 0.08 mole of Na₂SO₄ out of 0.09 mole given is neede to react completely with 0.08 mole of BaCl₂.

Therefore, BaCl₂ is the limiting reactant

How to determine the mass of the precipitate (BaSO₄) formed

Balanced equation

Na₂SO₄(aq) + BaCl₂(aq) —> BaSO₄(s) + 2NaCl(aq)

From the balanced equation above,

1 mole of BaCl₂ reacted to produce 1 mole of BaSO₄.

Therefore,

0.08 mole of BaCl₂ will also react to produce 0.08 mole of BaSO₄.

The mass of BaSO₄ can be obtained as follow

  • Mole of BaSO₄ = 0.08 mole
  • Molar mass of BaSO₄ = 137 + 32 + (16×4) = 233 g/mol
  • Mass of BaSO₄ =?

Mass = mole × molar mass

Mass of BaSO₄ = 0.08 × 233

Mass of BaSO₄ = 18.64 g

Learn more about stoichiometry:

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