chemistry

The combustion of 1.210 g of a mixture of CH4 and C2H6 gives 3.372 g of CO2

I have to determine the composition of the mixture (mastic %)

and

molar fraction of CH4 in the mixture
CH4 + C2H6: CO2 + 2H2O

C2H6 + O2: CO2 + H2O

CH4 + C2H6 + 02: CO2 +H2O

Thanks for the help

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asked by Ele
  1. This is a problem with two unknowns; therefore, you need two equations and solve them simultaneously. Write and balance the combustion equations.
    CH4 + 2O2 ==> CO2 + 2H2O
    2C2H6 + 7O2 ==> 4CO2 + 6H2O

    Let X = grams CH4
    and Y = grams C2H6
    ----------------------
    eqn 1 is X + Y = 1.210

    For equation 2 you want to convert X to grams CO2 from the CH4 component and convert Y to grams CO2 from the C2H6 component.
    I will use mm for molar mass.
    X(mm CO2/mmCH4) + Y(4*mm CO2/2*mm C2H6) = 3.372

    Solve those two equations simultaneously for X and Y to give grams each. I don't know what mastic % means; I assume you want % by mass.
    %CH4 = (X/1.210)*100 = ?
    %C2H6 = (Y/1.210)*100 = ?

    I assume you can handle the mole fraction part. Post your work if you get stuck.

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  2. Can you give me the result because mine is a negative mass

    Thank you

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    posted by Ele
  3. I don't get a negative number. Post your work and I'll find the error.

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  4. Thank you

    I post my work but there is something wrong

    For the CO2: M=44
    For CH4 M= 16
    For C2H6 M= 30

    number of mole of CO2= 3,372/44 = 0,766

    Could you explain me why you put: Y(4xmm CO2/2mmC2H6)?

    I would like to use

    m1/M1 + m2/M2= O,766
    and m1+m2= 1,210

    but don't know how to manage with C2H6 in relation with CO2

    Thank you very much for you attention

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    posted by Ele
  5. You need to re-read my explanation. If you want to convert 1.210 g to mols that's ok but more work than you need to do.
    I presume m1 is the mass of CH4 (that's my X) and m2 is the mass of C2H6(that's my Y). Your equation of
    m1/M1 + m2/M2 = 0.766 is right but unnecessary.
    m1 + m2 = 1.210 is right, also, however, it is not a different equation. The first one is mole = moles and the second one is grams = grams. You can't use both of these. For the second equation you must convert grams CH4 (that's your m1) in the mixture to grams CO2 produced by the CH4. I did that above with X(44/16). That would be m1(44/16) in your scheme.
    Then you need to convert grams C2H6 (that's your m2) to grams CO2 produced from the C2H6. That y(88/30) in mine and m2(88/30) in yours. Then those added together is actually g CO2 produced by CH4 + grams CO2 produced by C2H6 = 3.372 g total.
    That second equation becomes for me x(44/16) + Y(88/30) = 3.72 and yours is
    m1(44/16)+ Y(88/30) = 3.372

    Solve those two equations for X and Y for mine or for m1 and m2 for yours.
    Post your work if you get stuck. I have effectively answered this twice, once with my symbols and once with yours. It shouldn't be a problem to use X and Y and not m1 and m2 or the reverse but use what you feel familiar with.


    You want to convert

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  6. thank you for your help. It is very useful
    I'll do it and post my work

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    posted by Ele
  7. My result is m1=0,96 gr and the pourcentage is 0,96/1,21O = 79% (for CH4)

    For m2 it is the difference: 21% (for C2H6)

    but... the result I should found is
    75,6% for m1 (for CH4)
    and 24,4 % for m2 (for C2H6)

    Thank you for your help

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    posted by Ele

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