View Full Version : air/fuel ratio question (urgent again)
Ok, I have an exam on monday and both my lecturers and tutor have gone walkabouts. So you guys are my last resort. Thanks. :)
I get given a combustion chemical equation.
Looks like this : aC6H18 + bO2 + 3.76bN2 --> 9.5CO + 9.5CO2 + 9.5H2O + 71.5N2
I solve for the coefficients.
I can find the gravimetric a/f ratio, but I have no idea what to do to find the volumetric a/f ratio.
Also, for the equivalence ratio, I have the actual ratio, so to find the stoich. ratio, do I just have to rewrite the equation so it looks like --> xCO2 + yH2O +zN2??
Thanks again guys.
Gokul43201
Nov5-05, 11:29 PM
Ok, I have an exam on monday and both my lecturers and tutor have gone walkabouts. So you guys are my last resort. Thanks. :)
I get given a combustion chemical equation.
Looks like this : aC6H18 + bO2 + 3.76bN2 --> 9.5CO + 9.5CO2 + 9.5H2O + 71.5N2
I solve for the coefficients. Looks like there's an error in the equation. For instance, solving for 'a' from Carbon gives 6a = 19, but solving from Hydrogen gives 18a = 19. This is clearly a contradiction. So, the given equation appears to be flawed...or is there something I'm missing ?
I can find the gravimetric a/f ratio, but I have no idea what to do to find the volumetric a/f ratio. Assuming you can correctly find a and b, you can calculate the molar a/f ratio (using the fact that 1 mole of O2 is found in about 4.76 moles of air). The molar ratio is the same as the volumetric ratio, because a mole of any gas occupies the same volume under the same conditions (temperature, pressure). Finding the gravimetric ratio requires multiplying by molar masses (molecular weights).
Also, for the equivalence ratio...How is the equivalence ratio defined ?
Im not sure if that equation is right, Im pretty sure, but not 100%.
I understand the vol. ratio now. Thanks :)
As for the equivalence ratio, its defined as actual ratio/stoich ratio.
But dont worry about trying to explain everything to me, I had the exam this morning, I think it went well. :)
Thanks for the help Gokul!
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