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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Answer 1 · 2018-03-09T06:26:49

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")#