Question

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"`?

Answer 1

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")`