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Be sure to answer all parts.A sample of natural gas contains 7.424 moles of methane (CH4), 0.691 moles of ethane (C2H6), and 0.243 moles of propane (C3H8). If the total pressure of the gases is 3.78 atm, what are the partial pressures of the gases

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Answer:

  • Partial pressure of methane = 3.36 atm
  • Partial pressure of ethane = 0.30 atm
  • Partial pressure of propane = 0.12 atm

Explanation:

To solve this problem we need to keep in mind that the partial pressure of a gas is equal to the mole fraction of the gas multiplied by the total pressure of the mixture.

First we calculate the mole fraction of each gas:

Total number of moles = 7.424 + 0.691 + 0.243 = 8.358 moles

  • Mole fraction of methane = 7.424/8.358 = 0.89
  • Mole fraction of ethane = 0.691/8.358 = 0.08
  • Mole fraction of propane = 0.243/8.358 = 0.03

Now we calculate the partial pressure of each gas:

  • Partial pressure of methane = 0.89 * 3.78 atm = 3.36 atm
  • Partial pressure of ethane = 0.08 * 3.78 atm = 0.30 atm
  • Partial pressure of propane = 0.03 * 3.78 atm = 0.12 atm

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