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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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Write the concentration equilibrium constant expression for this reaction.
Akimi4 [234]

In a chemical reaction, the equilibrium constant refers to the value of its reaction quotient at chemical equilibrium, that is, a condition attained by a dynamic chemical system after adequate time has passed, and at which its composition has no measurable capacity to undergo any kind of further modification.  

The given reaction is: HCN (aq) + OH⁻ = CN⁻ (aq) + H2O (l)

The equilibrium constant = product of concentration of products / product of concentration of reactants

(Here, H2O is not considered as its concentration is very high)

So, Keq = [CN⁻] / [HCN] [OH⁻]


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3 years ago
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From the given choices
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3 years ago
Read 2 more answers
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Answer:

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V=volume

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