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imqwerty

  • one year ago

By the reaction of of carbon & oxygen , a mixture of co & co2 is obtained. What is the composition by mass of the mixture obtained when 20grams of O2 reacts with 12grams of carbon??

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  1. anonymous
    • one year ago
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    Do you have the chemical formula for this reaction?

  2. imqwerty
    • one year ago
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    4C + 3O2 -> 2CO2 + 2CO

  3. anonymous
    • one year ago
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    So first you need to convert 20g of O2 into moles.

  4. anonymous
    • one year ago
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    And then the 12g of C into moles.

  5. imqwerty
    • one year ago
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    we get 1mole of C reacting with 0.625 moles of O2

  6. imqwerty
    • one year ago
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    i think that the limitating reagent concept apply here

  7. anonymous
    • one year ago
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    Yep that's it.

  8. anonymous
    • one year ago
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    Do you know how the calculation works?

  9. anonymous
    • one year ago
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    @imqwerty

  10. imqwerty
    • one year ago
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    1st i got the limitating reagent as O2 2nd - according to the equation 3 moles of O2 give 2moles of CO2 and 2 moles of CO. 3rd - we have 20/32 moles of O2 so we will get 5/12 moles of both co and co2 4th - we have the moles of co and co2 nw we calculate their mass 5th - mass of co = 35/3gm and mass of co2 = 55/3 6th mass of co2:mass of co = 11:7 but the answer given is 21:11

  11. anonymous
    • one year ago
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    How did you get that ratio?

  12. imqwerty
    • one year ago
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    mass of co2/mass of co = 55/3 x 3/35 =11:7

  13. anonymous
    • one year ago
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    It seems like what your doing is right, and that answer doesn't seem to be right to me, as they got 21g for CO2 nO2 = m/M = 20g/32g/mol = 0.625mol nCO2 = nO2 x 2/3 = 0.41667 mol mCO2 = M x m = 0.41667 mol x 44.01g/mol = 18.333 g = 55/3 They used the mass of O2 needed rather than the actual mass.

  14. anonymous
    • one year ago
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    My ratio 18.338967/11.671767 ~ 1.5714 thats almost the same as 11/7 = 1.5714 So its correct to me, but then again maybe u should double check with your teacher as to why the answer is different.

  15. imqwerty
    • one year ago
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    thanks @ShizukaTheOtaku

  16. anonymous
    • one year ago
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    No problem.

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