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 <em>moles of the limiting reactant</em> we <u>calculate how many moles of BaSO₄ would have been produced if the % yield was 100%</u>:
- 60.8 mmol K₂SO₄ *
= 60.8 mmol BaSO₄
Then we <u>convert that theoretical amount into grams</u>, using the <em>molar mass of BaSO₄</em>:
- 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%