Answer to Question #21611 in Organic Chemistry for mimi
ANS: I calculated this to be 16.7:1 (based on the fact that i need 12.5 O2 for on 1 of C8H18 and O2 is only 21% of air.
my question is the next part.
determine the excess air factor if a dry volumetric analysis of the products of the fuel is C02(9%) C0(2%) N2(89%).
C8H18 + 12.5O2 ---> 8CO2 + 9H2O , where CO2 is produced
In your case one of the products is CO. It means that O2 was in shortage, so if there is no O2 in products, but only CO2 (9%), CO (2%), N2 (89%), air was not in excess.
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