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Water from gasoline calc.

There has been some speculation or information or... whatever, about 
how much water will be produced when a gallon of gasoline is burned.  
Well, here are the calculations based upon some perhaps rather specious 

Assumptions:  1.  Gasoline is pure "isooctane" as in 100 octane fuel.  It 
isn't but, what the heck, I'm the one making the assumptions.  :-)  2.  
Gasoline's "formula" averages out to C8H18 (which it doesn't, but see #1)  
3.  Gasoline burns to form CO2 and water exclusively  (pretty close) 
according to the following equation (which neglects imperfect combustion 
which may lead to CO formation among other things).

            2 C8H18 + 25 O2  --->  16 CO2  +  18 H2O

One liter of "gasoline" will weigh 692 grams (density of "isooctane" is 
0.692 g/mL).  Weight of water formed = (692 * 18 * 18)/(2 * 114) = 983 
grams.  (Please note allofyouengineersoutthere that limited significant 
figures are used in the calculations.  The numbers are only as good as 
the assumptions.  :-) )

This corresponds to approximately 983 mL assuming water density = 1.000 
(which it isn't except at 15.5 degree C).

Thus one gallon of gasoline will produce about 0.983 gallon of water.  (A 
reasonable estimate of the actual which will undoubtedly vary some from 
the actual due to the sloppiness of the assumptions above.)  This is pretty 
close to the 1 to 1 estimate given earlier by (who was it?,  Bob Houk?
I forget) someone.

| Robert L. (Bob) Myers  <rmyers@olie.wvitcoe.wvnet.edu>     | 
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