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Vanyuwa [196]
3 years ago
11

Explain how it is possible for 10 grams of methane fuel to burn and emit 27 grams of carbon dioxide. Discuss whether or not this

reaction obeys the law of conservation of mass.
Chemistry
1 answer:
Rufina [12.5K]3 years ago
7 0

This reaction obeys the law of conservation of mass.

Explanation:

In the burning reaction, methane (CH₄) react with oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O):

CH₄ + 2 O₂ → CO₂ + 2 H₂O

Now we calculate the number of moles of methane and carbon dioxide:

number of moles = mass / molar weight

number of moles of methane = 10 / 16.04 = 0.62 moles

number of moles of carbon dioxide = 27 / 44.01 = 0.61 moles

From the chemical reaction we see that 1 mole of methane produces 1 moles of carbon dioxide so 0.6 moles of methane produces 0.6 mole of carbon dioxide. This reaction obeys the law of conservation of mass because the mass of reactants is equal to the mass of products.

Learn more about:

combustion reaction

brainly.com/question/14117102

#learnwithBrainly

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Explanation:

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Answer is: the molar mass od sodium carbonate (Na₂CO₃) is 106.0 g/mol.

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8 0
3 years ago
If 2.00 moles of H₂ and 1.55 moles of O₂ react how many moles of H₂O can be produced in the reaction below?
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Hope this helped! :^)

4 0
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