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ziro4ka [17]
4 years ago
9

How many molecules of carbon dioxide are created when 16 molecules of propane react in the following

Chemistry
1 answer:
Grace [21]4 years ago
5 0

Answer:

48 molecules of CO₂

Explanation:

I think you made a mistake in your question.  The formula for propane is C₃H₈, not C₃H₃.  But, I will give you the answer for both cases.

For C₃H₃:

First you have to balance the equation.

4 C₃H₃  +  15 O₂  ⇒  12 CO₂  +  6 H₂O

Next, you need to use the mole ratios between C₃H₃ and CO₂ to find the amount of molecules of CO₂ you will produce with the given amount of C₃H₃.

(16 mol's C₃H₃) × (12 mol's CO₂/4 mol's C₃H₃) = 48 mol's CO₂

You will get 48 molecules of CO₂.

For C₃H₈:

Balance the equation.

C₃H₈  +  5 O₂  ⇒  3 CO₂  +  4 H₂O

Use the mole ratios between C₃H₈ and CO₂.

(16 mol's C₃H₈) × (3 mol's CO₂/1 mol's C₃H₈) = 48 mol's CO₂

You will get 48 molecules of CO₂ for this equation as well.

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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 bari
BabaBlast [244]

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>:

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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
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Finally we calculate the % yield:

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3 years ago
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