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How many moles of methane gas are required when there are 7.55 grams of oxygen gas available? Please someone explain how to solve this problem, I would really appreciate it.

Sagot :

CintiaSalazar

Answer:

0.236 moles of methane gas are required when there are 7.55 grams of oxygen gas available.

Explanation:

The balanced reaction is

CH₄ + 2 O₂ → CO₂ + 2 H₂O

By reaction stoichiometry (that is, the ratio of reactants and products in the reaction), the following amounts of moles of each compound participate in the reaction:

  • CH₄: 1 mole
  • O₂: 2 moles
  • CO₂: 1 mole
  • H₂O: 2 moles

Being the molar mass of each compound:

  • CH₄: 16 g/mole
  • O₂: 32 g/mole
  • CO₂: 44 g/mole
  • H₂O: 18 g/mole

Then, by reaction stoichiometry, the following mass amounts of each compound participate in the reaction:

  • CH₄: 1 moles* 16 g/mole= 16 grams
  • O₂: 2 moles* 32 g/mole= 64 grams
  • CO₂: 1 mole* 44 g/mole= 44 grams
  • H₂O: 2 moles* 18 g/mole= 36 grams

So, you can apply the following rule of three: if by reaction stoichiometry 64 grams of oxygen react with 2 moles of methane, 7.55 grams of oxygen react with how many moles of methane?

[tex]moles of methane=\frac{7.55 grams of oxygen*2 moles of methane}{64 grams of oxygen}[/tex]

moles of methane= 0.236

0.236 moles of methane gas are required when there are 7.55 grams of oxygen gas available.

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