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1 July, 12:16

A sample of propane (C3H8) is placed in a closed vessel together with an amount of O2 that is 3 times the amount needed to completely oxidize the propane to CO2 and H2O at constant temperature. Calculate the mole fraction of each component in the resulting mixture after oxidation assuming that the H2O is present as a gas, and the partial pressure of each gas assuming that the total pressure inside the vessel following the reaction is 5 atm. Assume all gases are ideal

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  1. 1 July, 12:43
    0
    Mol fraction CO2 = 0.176

    Mol fraction H2O = 0.235

    Mol fraction O2 = 0.588

    Partial pressure of CO2 = 0.88 atm

    Partial pressure of H2O = 1.175 atm

    Partial pressure of O2 = 2.94 atm

    Explanation:

    Step 1: Data given

    amount of O2 = 3 times the amount needed to completely oxidize propane

    Step 2: The balanced equation:

    C3H8 + 5 O2 → 3 CO2 + 4 H2O

    Step 3: Calculate moles at the start

    Suppose there is 1.00 mol of C2H8 to start.

    Then there is 1.00 * 5 * 3.00 = 15.0 moles of O2 to start.

    Step 4: Calculate moles at the equilibrium

    After the reaction, 1.00 mol of C3H8 is completely consumed

    There is 3*1.00 = 3.00 moles of CO2 and 4*1.00 = 4.00 moles H2O moles produced

    There will be consumed 5*1.00 = 5.00 moles of O2

    There will remain 15.00 - 5.00 = 10.00 moles O2

    Step 5: Calculate total moles of gases at the end

    Total moles of gases at the end: 3 moles CO2 + 4 moles H2O + 10 moles O2 = 17.0 moles

    Step 6: Calculate mol fraction

    Mol fraction CO2 = 3.00 moles / 17.00 moles = 0.176

    Mol fraction H2O = 4.00 moles / 17.00 moles = 0.235

    Mol fraction O2 = 10.00 moles / 17.00 moles = 0.588

    Mol fraction CO2 + mol fraction H2O + mol fraction O2 = 0.176 + 0.235 + 0.588 = 1

    Step 7: Calculate partial pressure

    Partial pressure of CO2 = 0.176 * 5.00 atm = 0.88 atm

    Partial pressure of H2O = 0.235 * 5.00 atm = 1.175 atm

    Partial pressure of O2 = 0.588 * 5.00 atm = 2.94 atm

    Partial pressure of CO2 + partial pressure of H2O + partial pressure of O2 = 0.88 atm + 1.175 atm + 2.94 atm = 5 atm
  2. 1 July, 15:19
    0
    final mole fraction of O₂ = 58.84%, CO₂ = 17.64%, H₂O = 23.52%.

    final partial pressure of O₂ = 2.942 atm, CO₂ = 0.882 atm, H₂O = 1.176 atm.

    Explanation:

    Assuming that propane is present as a gas, and also that the combustion is complete:

    C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

    then taking as a reference propane sample = 1 mol, then

    initial moles of O₂ = 3 * 5 moles = 15 moles

    final moles of O₂ = 3 * 5 moles - 5 moles = 10 moles

    final moles of CO₂ = 3 moles

    final moles of H₂O = 4 moles

    total number of moles = 10 + 3 + 4 = 17 moles

    final mole fraction of O₂ = 10/17 = 0.5884 = 58.84%

    final mole fraction of CO₂ = 3/17 = 0.1764 = 17.64%

    final mole fraction of H₂O = 4/17 = 0.2352 = 23.52%

    From Dalton's law for ideal gases, the partial pressure p=P*x then

    final partial pressure of O₂ = 5 atm * 10/17 = 2.942 atm

    final partial pressure of CO₂ = 5 atm * 3/17 = 0.882 atm

    final partial pressure of H₂O = 5 atm * 1/17 = 1.176 atm
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