Question

When 80.0 mL of a 0.812 M barium chloride solution is combined with 40 mL of a 1.52 M potassium sulfate solution, 10.8 g of barium sulfate precipitates. What is the % yield of this reaction

Answer 1

Answer:

76.1%

Explanation:

The reaction that takes place is:

• BaCl₂ + K₂SO₄ → BaSO₄ + 2KCl

First we determine how many moles of each reactant were added:

• BaCl₂ ⇒ 80 mL * 0.812 M = 64.96 mmol BaCl₂

• K₂SO₄ ⇒ 40 mL * 1.52 M = 60.8 mmol K₂SO₄

Thus K₂SO₄ is the limiting reactant.

Using the moles of the limiting reactant we calculate how many moles of BaSO₄ would have been produced if the % yield was 100%:

• 60.8 mmol K₂SO₄ * $\frac{1mmolBaSO_4}{1mmolK_2SO_4}$ = 60.8 mmol BaSO₄

Then we convert that theoretical amount into grams, using the molar mass of BaSO₄:

• 60.8 mmol BaSO₄ * 233.38 mg/mmol = 14189.504 mg BaSO₄

• 14189.504 mg BaSO₄ / 1000 = 14.2 g BaSO₄

Finally we calculate the % yield:

• % yield = 10.8 g / 14.2 g * 100 %

• % yield = 76.1%