wrote in message
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"Jeffrey Lebowski" wrote:


"Steve Spence" wrote in message
...

If a propane burner is emitting CO, then it's not adjusted properly.
Our
gas company measured 13 ppm using the air-free method.


http://en.wikipedia.org/wiki/Combustion


quoting from that source.....

Chemical equation

Generally, the chemical equation for stoichiometric burning of
hydrocarbon in oxygen is as follows:

C_xH_y + (x + \frac{y}{4})O_2 \rightarrow \; xCO_2 +
(\frac{y}{2})H_2O

For example, the burning of propane is:

C_3H_8 + 5O_2 \rightarrow \; 3CO_2 + 4H_2O

The simple word equation for the combustion of a hydrocarbon in oxygen
is:

\textrm{Fuel} + \textrm{Oxygen} \rightarrow \; \textrm{Heat} +
\textrm{Water} + \textrm{Carbon\ dioxide}

If the combustion takes place using air as the oxygen source, the
corresponding equations a

C_xH_y + (x+ \frac{y}{4})O_2 + 3.76(x+ \frac{y}{4})N_2 \rightarrow
\; xCO_2 + (\frac{y}{2})H_2O + 3.76(x + \frac{y}{4})N_2

For example, the burning of propane is:

C_3H_8 + 5O_2 + 18.8N_2 \rightarrow \; 3CO_2 + 4H_2O + 18.8N_2

The simple word equation for the combustion of a hydrocarbon in air
is:

\textrm{Fuel} + \textrm{Air} \rightarrow \; \textrm{Heat} +
\textrm{Water} + \textrm{Carbon\ dioxide} + \textrm{Nitrogen}


bottom line, I couldnt see where CO by itself is produced.......by
combusting propane and oxygen or propane and air..... co2, yes, co,
no.

Fish,

Yes I know this--here's where it does get produced :

http://en.wikipedia.org/wiki/Combustion#Incomplete

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