Using the chemical equation below, determine the energy released by burning 2 moles of propane, #"C"_3"H"_8#? a) 4438.4 kJ b) 11096 kJ c) 2219.2 kJ d) 1109.6 kJ

Using values from Thermodynamic Properties table and applying to the Hess's Law Equation, the following was obtained ...

The problem provides you with the thermochemical equation that describes the combustion of propane.

#"C"_ 3"H"_ (8(g)) + 5"O"_ (2(g)) -> 3"CO"_ (2(g)) + 4"H"_ 2"O"_ ((l)) + color(blue)("2219.2 kJ")#

This tells you that when #1# mole of propane undergoes combustion, #color(blue)("2219.2 kJ")# of heat are being given off.

In your case, you want to know how much heat will be given off when #2# moles of propane undergo combustion. You can use the aforementioned ratio as a conversion factor to get

#2 color(red)(cancel(color(black)("moles C"_3"H"_8))) * overbrace("2219.2 kJ"/(1color(red)(cancel(color(black)("mole C"_3"H"_8)))))^(color(red)("given by the balanced thermochemical equation")) = color(darkgreen)(ul(color(black)("4438.4 kJ")))#

Alternatively, you can multiply all the products and all the reactants, including the heat given off by the reaction, by #color(red)(2)#

#color(red)(2)"C"_ 3"H"_ (8(g)) + (color(red)(2) * 5)"O"_ (2(g)) -> (color(red)(2) * 3)"CO"_ (2(g)) + (color(red)(2) * 4)"H"_ 2"O"_ ((l)) + (color(red)(2) * color(blue)("2219.2 kJ"))#

to get

#2"C"_ 3"H"_ (8(g)) + 10"O"_ (2(g)) -> 6"CO"_ (2(g)) + 8"H"_ 2"O"_ ((l)) + color(darkgreen)(ul(color(black)("4438.4 kJ")))#

Once again, you can say that when #2# moles of propane undergo combustion, #"4438.4 kJ"# of heat are being given off.