In article ,
Edwin Pawlowski wrote:
Mike in Mystic wrote:
Hi Ed,

The oxygen in the water isn't coming from the propane, it's coming
from the air. 2 of the mass units in water are from propane, but 16
(the oxygen) are from air. If you work it out stochiometrically it
makes sense.
Mike

"Edwin Pawlowski" wrote in message
. com...
Robert Bonomi wrote:
It's all in how you measure things --

One lb of propane, when burned, creates approximately 2-1/4 lbs of
water (in vapor form, of course.) This is roughly 35 fl.oz.

OK, but Robert says it "creates" water in vapor form. If it is in the air,
it is vapor and does not have to be created. Raising the temperature changes
the relative humidity and the dew point and thus the condensation. If there
is water in the propane it can be released to the air in the form of vapor.
Or a chemcial reaction can change one physical item to another, such as milk
into cheese. I just don't know of any mass that can be changed to another
form and make a greater mass though.

Robert knows _generally_ knows what he's talking about.

I'm sure of that -- he's me. grin

In this case, I _do_ have the theory right, but was giving propane credit for
being about 30% lighter than it actually is -- correct figures below.


This is "Basic Chemistry". What's going on is the 'chemical reaction' you
alluded to.

"Created" _is_ the correct term.

Propane *doesn't* burn "all by itself".

One has to have oxygen present as well. Whether there is _any_ water vapor
present in the 'air' is _irrelevant_. OXYGEN, on the other hand, is critical.

The chemical reaction:

(1) C3H8 + (5) O2 = (3) CO2 + (4) H2O(gas) + (1996.04 BTU/mole){heat}
or
"one molecule of propane, together with five molecules of oxygen produces
three molecules of carbon-dioxide and four molecules of water and liberates
'1996.04/(6.022*10^23) BTU' heat in the process"
{ Note: 1.552 oz (44 grams) by weight, of propane when burned completely,
gives off approx 1996.04 BTUs of heat -- an additional 4752 BTUs can be
liberated if all the created water vapor is reduced to liquid.)


Due to the differing atomic weights, things work out (by *mass*) as:

1-3/8 lb propane + 5 lb oxygen = 4-1/8 lb CO2 + 2-1/4 lb water

As should be obvious,this is "6-3/8 lbs in" and "6-3/8 lbs out".

Mass *is* preserved; you just have to account for _everything_ in the reaction.


Of the 2-1/4 lbs of water 'created', only 1/4 lb of that total weight comes
from components of the propane. The other 2 lbs comes from the oxygen in
the air. The remaining 1-1/8 lb of the weight of the propane, in conjunction
with 3 lbs of oxygen in the air, turns into 4-1/8 lbs of carbon-dioxide.

Regardless of however much water vapor _was_ in the air before you lit the
fire, after burning 1-3/8 pounds of propane, there is an *additional* 2-1/4
lbs of water "somewhere", either in the air as water vapor, or having
'condensed out' on something. (like the tables of a cold jointer




A final note: "This material _will_ be on the exam on Friday!"