How do I do this? DeltaH_(rxn) = -"2219.9 kJ/mol propane", so how much heat is released by combustion of "16.50 kg propane"?

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1 Answer
Mar 9, 2018

The heat released is 8.306 xx 10^5 "kJ". Why is this supposed to be positive even though heat is released?


Based on what you got in part 1+2, you got DeltaH_(rxn) = -"2219.9 kJ/mol propane", for the reaction:

"C"_3"H"_8(g) + 5"O"_2 -> 3"CO"_2(g) + 4"H"_2"O"(g)

Knowing DeltaH_(rxn), we just have

"16.50 kg C"_3"H"_8 = "16500 g C"_3"H"_8

=> 16500 cancel"g" xx "1 mol"/(44.1 cancel("g C"_3"H"_8)) = "374.15 mols C"_3"H"_8

Therefore:

color(blue)(q_(rxn)) = -"2219.9 kJ"/cancel("mol C"_3"H"_8) xx 374.15 cancel("mols C"_3"H"_8) = "830574.83 kJ"

= -color(blue)(8.306 xx 10^5 "kJ")