Results 1 to 2 of 2

Thread: Air Fuel Ratio's

  1. #1 Air Fuel Ratio's 
    New Member
    Join Date
    Jan 2009
    Hi there,

    wondered whether anyone might be able to help me out with my little problem

    I have been given the chemical formula C7 H16. I have to create a balanced equation for the burning of the above with 30% excess air.

    In the end I ended up with:

    C7H16 + (1.3 x 11O2 + 79/21N2) - - ~ 7CO2 + 8H20 + 79/21N2 + 3.3O2[/size]

    ...Which I think/hope is right...

    anyway im now asked to find the 'air/ fuel ratio by mass', so I divide the total for oxegen on the left hand side (457.6) and the total for nitrogen (105.33) by the total for fuel -C7H16 (100) and it gives me 5.277kg/kg. Have i done this right?

    then I am asked to find'percentage analysis by mass of product'. to do this I have taken all the elements from the left hand side of the equation and divided them by 662.93 (the total mass for left hand side) to give me a percentages -

    C7 - 12.7%
    H16 - 2.4%
    11O2 - 69%
    79/21N2 - 15.9%

    The last thing I am asked to do is to show a 'percentage volume of oxygen in dry products of combustion'.

    Anyone any ideas on how I can show this? I dont quite understand the question.

    Please help me.

    Many Thanks

    Stuart Firth
    Reply With Quote  


  3. #2  
    Forum Masters Degree organic god's Avatar
    Join Date
    Feb 2008
    dry products will be everything on the right hand side but water.

    everything is mathematical.
    Reply With Quote  

Posting Permissions
  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts