Given that DeltaH_"combustion"^@ for "propane" is -2043*kJ*mol^-1, what mass of propane is required to give 425*kJ?

1 Answer
Nov 9, 2015

You have the equation; treat the energy as a product!

Explanation:

C_3H_8(g) + 5O_2(g) rarr 3CO_2(g) + 4H_2O(g) + 2043 kJ

Normally, we would write the energy given as a negative quantity. Here, I have it listed as a reaction product (as indeed it is). Combustion of 1 mol propane gives 2043 kJ of energy. (That is 2043 kJ of energy are evolved per mole of reaction as written!)

So if 425 kJ of energy are evolved, then (425*kJ)/(2043*kJ) ~= 1/5. Thus, approximately, 1/5 of a mole of propane must be combusted with stoichiometric oxygen to evolve this amount of energy. You can calculate the molar quantity of propane in grams, and you need approx. 1/5 of this.