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    Hi

    Can anyone help with this question?

    - Write a balanced equation for the complete combustion of octane in oxygen?

    Thanks...


    :tsr2:
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    (Original post by seaspray)
    Hi

    Can anyone help with this question?

    - Write a balanced equation for the complete combustion of octane in oxygen?

    Thanks...


    :tsr2:
    CnH2n+2 + (1.5n+0.5)O2 ----> nCO2 + (n+1)H2O

    Therefore with n = 8

    C8H18 + 12.5O2 ----> 8CO2 + 9H2O
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    (Original post by Widowmaker)
    CnH2n+2 + (1.5n+0.5)O2 ----> nCO2 + (n+1)H2O

    Therefore with n = 8

    C8H18 + 12.5O2 ----> 8CO2 + 9H2O
    Thank you so much.

    How do I use that equation to calculate the volume of air (containing 20% Oxygen) needed to burn one mole of octane?

    Thanks again
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    (Original post by seaspray)
    Thank you so much.

    How do I use that equation to calculate the volume of air (containing 20% Oxygen) needed to burn one mole of octane?

    Thanks again
    Mole ratio
    C8H18:O2
    = 1:12.5

    Therefore 12.5 moles of oxygen required.

    But air is 20% oxygen, so 5 x 12.5 moles of air = 60 moles

    1 mole of gas occupies 24dm^3 under standard conditions.

    So volume of air required = 60 x 24dm^3 = 1440dm^3
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    (Original post by Widowmaker)
    Mole ratio
    C8H18:O2
    = 1:12.5

    Therefore 12.5 moles of oxygen required.

    But air is 20% oxygen, so 5 x 12.5 moles of air = 60 moles

    1 mole of gas occupies 24dm^3 under standard conditions.

    So volume of air required = 60 x 24dm^3 = 1440dm^3
    Thanks again, but can you explain something for me?.....

    Air = 20% oxygen, but where do you get the 5 from to give 60 moles?

    Thanks.
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    (Original post by seaspray)
    Thanks again, but can you explain something for me?.....

    Air = 20% oxygen, but where do you get the 5 from to give 60 moles?

    Thanks.
    Air consists of 20% oxygen so you need 5 times the volume of oxygen in air to give you 12.5 moles of oxygen. I.e. 12.5 moles/60 moles x 100% = 20%. If air was 100% oxygen you'd just use 12.5 moles. I.e. 12.5/12.5 x 100% = 100% (if air contained 100% oxygen)
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    (Original post by Widowmaker)
    Air consists of 20% oxygen so you need 5 times the volume of oxygen in air to give you 12.5 moles of oxygen. I.e. 12.5 moles/60 moles x 100% = 20%. If air was 100% oxygen you'd just use 12.5 moles. I.e. 12.5/12.5 x 100% = 100% (if air contained 100% oxygen)
    Thank you....it's so obvious now you explained

    There are two more parts to the question...

    - Burning 1g of Octane produces 48.4 kJ of energy. How much heat energy is produced by burning 1g of Methane?

    - Burning 1dm3 of liquid octane releases 33800kJ of energy. How much heat energy is released on burning 1dm3 of gaseous methane?

    I don't even know where to start on these two

    Thank you so much for your help
 
 
 
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