I have a question about incomplete combustions. I think that you can have carbon monoxide, carbon dioxide, water and carbon or some combination of these in the products of incomplete combustions. Is this right? For Pentane, would it be correct to show one possible equation as 2C5H12+13O2--12H2O+6O2+2C0+2C? thanks!

Question

I have a question about incomplete combustions. I think that you can have carbon monoxide, carbon dioxide, water and carbon or some combination of these in the products of incomplete combustions. Is this right? For Pentane, would it be correct to show one possible equation as 2C5H12+13O2-->12H2O+6O2+2C0+2C? thanks!

DrBob222 · Answer

Under controlled conditions I suppose it may be possible to have combinations; however, it is highly unlikely to result in the equation you wrote. First, it isn't balanced; second you have oxygen showing on BOTH sides of the equation. It is likely that CO and CO2 + H2O could co-exist as products and they do in various circumstances but I think uncombusted carbon is not likely. If there is enough oxygen present to break the C-H bonds to produce water then it is most likely that the C would be oxidized to CO at least and to CO2/CO mixture depending upon the amount of O2.

Anne · Answer

Thanks. Is there any way to tackle these incomplete combustion equations, any form to follow? I'd really appreciate it since I'm clearly on the wrong track.

DrBob222 · Answer

It's difficult to know how to answer a question that is so broad. Tackle for what? What do you want to accomplish? Here is the CO equation under incomplete combustion.
2C5H12 + 11 O2 ==> 10CO + 12H2O

Under complete combustion:
C5H12 + 8O2 ==> 5CO2 + 6H2O

Anne · Answer

Sorry for pestering you on this but I really appreciate your help. The way I first did the equation, I had too many products. My question is when I do these should the products only be in the form of either a)CO and H2O or B)C and H2O? So if I set up the products to have either of these combinations and then I balance will I have done it properly. thanks again.

GK · Answer

You can have three different chemical equations going on at the same time:
C5H12 + 8O2 ==> 5CO2 + 6H2O
2C5H12 + 11 O2 ==> 10CO + 12H2O
C5H12 + 3O2 ==> 5C + 6H2O
The relative amounts of C5H12 undergoing each reaction simultaneously varies depending on the availability of O2 and and rate of mixing of C5H12 with O2.

Anne · Answer

Thanks so much!