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Taya2010 [7]
3 years ago
8

Mr. Holmes and Mrs. Nogaki are having a good old fashioned Baby Back Ribs BBQ contest and are arguing over who’s BBQ produces le

ss greenhouse gasses. Mrs. Nogaki uses natural gas to power her BBQ and provided the balanced equation to Mr. Holmes as proof of how efficient her grill is.
CH4 + 2O2→CO2 +2H2O

Mrs. Nogaki claims that for every mole of natural gas burned, only 1 mole of CO2 is produced.

Mr. Holmes doesn’t have a prepared balanced equation but knows that the chemical formula of propane is C3H8 and that it is a combustion reaction.

Who is right?

Question 19 options:

Mrs. Nogaki is right because Mr. Holmes’s BBQ produces 2x less CO2 for each mole of fuel burned.


Both are right because in a combustion reaction, CO2 and water are always produced.


Mr. Holmes is right because his BBQ produces less CO2 for each mole of fuel burned than Mrs. Nogaki’s grill.


Mrs. Nogaki is right because Mr. Holmes’s BBQ produces 3x more CO2 for each mole of fuel burned.
Chemistry
1 answer:
ExtremeBDS [4]3 years ago
5 0

Answer:

Mrs. Nogaki is right because Mr. Holmes’s BBQ produces 3x more CO2 for each mole of fuel burned.

Explanation:

Now Mrs. Nogaki has already figured out the chemical combustion reaction behind the operation of her BBQ. It is pertinent to reproduce it here.

CH4(g) + 2O2(g)→CO2(g) +2H2O(g)

She already has this figured out but Mr. Holmes doesn't have any chemical reaction equation to back his claims. Let us help him with the correct combustion equation for propane.

C3H8(g) + 5O2(g) + 3CO2(g) + 4H2O(g)

We can clearly see from the reaction equation that Mr. Holmes BBQ produces three times more carbon IV oxide than Mr. Nogaki's BBQ so Mr. Nogaki was right in her claim after all.

Hence the answer!

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What mass, in grams, of CO2 and H2O<br> is formed from 2.55 mol of propane?
oksian1 [2.3K]

Answer:

336.6 grams of CO₂ and 183.6 grams of H₂O are formed from 2.55 moles of propane.

Explanation:

In this case, the balanced reaction is:

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reactant and product participate in the reaction:

  • C₃H₈: 1 mole
  • O₂: 5 moles
  • CO₂: 3 moles
  • H₂O: 4 moles

Being the molar mass of each compound:

  • C₃H₈: 44 g/mole
  • O₂: 16 g/mole
  • CO₂: 44 g/mole
  • H₂O: 18 g/mole

Then, by stoichiometry, the following quantities of mass participate in the reaction:

  • C₃H₈: 1 mole* 44 g/mole= 44 grams
  • O₂: 5 moles* 16 g/mole= 80 grams
  • CO₂: 3 moles* 44 g/mole= 132 grams
  • H₂O: 4 moles* 18 g/mole= 72 grams

So you can apply the following rules of three:

  • If by stoichiometry 1 mole of C₃H₈ forms 132 grams of CO₂, 2.55 moles of C₃H₈ how much mass of CO₂ will it form?

mass of CO_{2} =\frac{2.55 moles of C_{3} H_{8}*132 gramsof CO_{2} }{ 1 mole of C_{3} H_{8}}

mass of CO₂= 336.6 grams

  • If by stoichiometry 1 mole of C₃H₈ forms 72 grams of H₂O, 2.55 moles of C₃H₈ how much mass of H₂O will it form?

mass of H_{2}O =\frac{2.55 moles of C_{3} H_{8}*72 gramsof H_{2}O }{ 1 mole of C_{3} H_{8}}

mass of H₂O= 183.6 grams

<u><em>336.6 grams of CO₂ and 183.6 grams of H₂O are formed from 2.55 moles of propane.</em></u>

3 0
3 years ago
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