The balanced equation below shows the combustion of propane.
C3H8+5O2–>3CO2+4H2O+energy
A) How do I determine the total number of moles of CO2 produced during the test?

B) What is the formula for the primary metal hydroxide used in the lantern and how do i find it?

thanks in advance


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By the way, the reaction for the combustion of propane is C3H8(l) + 5O2(g) ? 3CO2(g) + 4H2O(g).

And, for that equation ?H = -2220 kJ/mol.

I greatly appreciate any help!


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Here in the Philippines LPG is contains 70% butane and 30% propane. How about if butane becomes only 40% and propane only 60% does it not affect the combustion process of a car engine?


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It is being proposed that the natural gas fired boiler be converted to LPG, for the sake of wobe’s index LPG will be mixed with air, how will the combustion process effected?


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This is the completed (as far as I plan on completing it) propane forced air heater. The fan is from a kitchen exhaust vent hood. The rest is just various pieces of 6″ and 8″ ductwork and 1/4 galvanized piping. The final piece to get the flame right was the polyurethane can in the back-center of the inner tube. In this run I estimate it to be putting out around 100000 BTU, but by reducing the amount of air bypass I can get it up to at least 200000 BTU, limited only by how much gas the 20 lb. cylinder will put out. For those interested I made a forum thread covering the process. backyardmetalcasting.com