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

1 Answer
anor277
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.

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