Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Reaction Formula

  1. Oct 13, 2004 #1
    Hi

    I have this reaction formula

    [itex]
    CH_{4} + 2O_{2} \rightarrow 2H_{2} O + CO_{2}
    [/itex]

    CH4 and O2 are contained in a canister which has a volume V = 500 Liters.

    The mass of CH4 is 100 grams and O2 has a mass of 100 grams.

    I have two questions:

    First if I want to calculate the mass of the generated [itex]CO_2[/itex].

    Can't this be done by the following calculation

    [itex]
    n(CH_4) = n(CO_2) \ \rightarrow \ n(CO_2) = \frac{m(CH_4)}{M(CH_4)} = ??
    [/itex]

    And then multiply it with [itex]M(CO_2)[/itex] ??

    Second: if the temperature in the canister is messured to be 105 degrees celcius at the end of the above reaction. How do I calculate the partial pressure of [itex]CH_{4}[/itex] at this temperature ??

    Many Thanks in advance :)

    Sincerely Fred
    Denmark
     
    Last edited: Oct 14, 2004
  2. jcsd
  3. Oct 14, 2004 #2
    CH4+ 202 --> 2H2O + CO2

    CH4 = 100g

    2O2 = 100g

    Mr of substances (one mole weighs):

    CH4 = 16.043g

    O2 = 31.999 so 2O2 = 63.998

    H2O = 18.015 so 2H2O = 63.030

    CO2 = 44.010.

    Now calculate number of moles:

    Number of moles = Mr / w (w = mass of substance)

    CH4 = 100 / 16.043 = 6.23 mol.

    02 = 100 / 69.998 = 1.43 mole.

    Mole of CO2 = 6.23 x 1/1.43 = 4.40 mol

    Mass = 4.4 x 44.010 = 193.644g.


    Part 2:

    Temperature = 105 + 273.15 = 378.15K

    Volume = 500 dm^3

    Now from previous answers:

    pO2 = 6.23(0.0821 (375.15)) / 500 = 0.68 atm.

    Wolfson.
     
  4. Oct 14, 2004 #3
    Hi

    Thanks for Your answer.

    I have about Your calculations.

    How is possible that the partial pressure of O2 molecules at the end of the reaction are the same as the p.pressure of CH4 molecules?

    Sincerely
    Fred

     
  5. Oct 16, 2004 #4
    from the mole fraction of CH4
     
  6. Oct 17, 2004 #5
    Sincerely

    /Fred
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Reaction Formula
  1. Possible Reaction (Replies: 1)

  2. Oxidation reaction (Replies: 9)

  3. Rate of reaction (Replies: 7)

Loading...