Express the pictured problem as a chemical equation.
2 answers:
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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%
Answer:
5.95g
Explanation:
1 = 1000 mL
∴ 100 mL = 100 ÷ 1000 = 0.1
Volume = <u>0.1 </u><u />
Concentration = <u>0.5 M</u>
Concentration =
0.5 =
No. of moles = 0.5 x 0.1 = <u>0.05 moles</u>
No. of moles =
Mass in 1 mole of KBr = 39 + 80 = 119g (39 is the mass of potassium and 80 is the mass of bromine)
0.05 =
x = 119 × 0.05 = 5.95g