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Hydrocarbons analysis

  1. Aug 8, 2010 #1
    1. The problem statement, all variables and given/known data
    10mLCxHy reacts with excess O2
    After reacting and cooling until 298K the volume GAS = 55mL. Then NaOH is added and then the final volume of GAS = 35mL. Give the emperical formular of the hydrocarbon used in this reaction.



    2. Relevant equations

    -CxHy + (X+Y/4)O2 -> X CO2 + Y/2 H2O

    -CO2+ 2NaOH -> Na2CO3 + H20


    3. The attempt at a solution
    10ml CxHY in the beginning and 55ml CO2 at the end => 1/X=10/55
    55ml/X=2*VolumeH20/Y
     
  2. jcsd
  3. Aug 8, 2010 #2

    Borek

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    Why do you think there was 55 mL of CO2?
     
    Last edited by a moderator: Aug 13, 2013
  4. Aug 8, 2010 #3
    Because at the end of the reaction after cooling there is 55mL gas, H20 is liquid then and I think you shouldn't take acount of the O2 in the 55mL
     
  5. Aug 8, 2010 #4

    Borek

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    So what was the gas adsorbed by the NaOH?
     
    Last edited by a moderator: Aug 13, 2013
  6. Aug 8, 2010 #5
    You are right, I meant the amount of CO2 after the first reaction.
     
  7. Aug 8, 2010 #6

    Borek

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    I think you are still misunderstanding the question. What gases were present in 55 mL, what gas is present in the remaining 35 mL?
     
    Last edited by a moderator: Aug 13, 2013
  8. Aug 8, 2010 #7
    I think the answer is CO2 for both questions
     
  9. Aug 8, 2010 #8

    Borek

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    So why part of the CO2 was adsorbed and part not?
     
    Last edited by a moderator: Aug 13, 2013
  10. Aug 8, 2010 #9
    20 mL absorbed of the CO2 created by the first reacting absorbed and 35 mL remaining after the second reaction
     
  11. Aug 8, 2010 #10

    Borek

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    What second reaction? There was just a one reaction producing carbon dioxide, and that was combustion.
     
    Last edited by a moderator: Aug 13, 2013
  12. Aug 8, 2010 #11
    NaOH reacts with CO2 (I guess)

    -CO2+ 2NaOH -> Na2CO3 + H20
     
  13. Aug 8, 2010 #12

    Borek

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    So how come some of CO2 was left?
     
    Last edited by a moderator: Aug 13, 2013
  14. Aug 9, 2010 #13
    Because there wasn't enough NaOH, or because there is a dynamic equilibrium. I am not sure what kind of reaction this is (<=> or ->)
     
  15. Aug 9, 2010 #14

    Borek

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    No. All CO2 was absorbed. What other gas was present in the mixture? Reread the question.
     
    Last edited by a moderator: Aug 13, 2013
  16. Aug 9, 2010 #15
    I see, so the gass at the end is the excess O2, this would mean that the volume CO2 after the first reaction = 20mL

    Molfraction=volumefraction for ideal gasses so if I am not mistaken :

    20mL/X=2*VolumeH20/Y=10mL/1=VolumeO2/(X+Y/4)

    But I still don't get the formular out of this.
     
  17. Aug 9, 2010 #16

    Borek

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    From ideal gas ratio of volumes is identical to ratio of moles, isnt't it? 10 mL of gas gave 20 mL of carbon dixode, how many carbon atoms per mole?

    Edit: not sure what to do with hydrogen, as far as I can tell there is no enough information. Or I have a senior moment :grumpy:
     
    Last edited by a moderator: Aug 13, 2013
  18. Aug 9, 2010 #17
    20mL/X=10mL so 2
     
  19. Aug 9, 2010 #18

    Borek

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    Correct. And as far as I can tell that's all that can be said.

    Unless I am missing something.

    Have you posted whole question?
     
    Last edited by a moderator: Aug 13, 2013
  20. Aug 9, 2010 #19
    I also think there is missing something, but that's the whole question.

    Thank you for the help anyway
     
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