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"?
1 Answer
Mar 9, 2018
The heat released is
Based on what you got in part 1+2, you got
"C"_3"H"_8(g) + 5"O"_2 -> 3"CO"_2(g) + 4"H"_2"O"(g)
Knowing
"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")