Question #75fed

The first thing that you need to do here is to determine the amount of heat that must be generated by the reaction in order to ensure that the pork roast absorbs #"1500 kJ"# of heat.

The problem tells you that the roast absorbs #14%# of the heat generated by the reaction, which means that for every #"100 kJ"# of heat given off by the barbeque, the roast absorbs #"14 kJ"#.

You can thus say that in order for the roast to absorb #"1500 kJ"# of heat, the barbeque must give off

#1500 color(red)(cancel(color(black)("kJ roast"))) * overbrace("100 kJ barbeque"/(14color(red)(cancel(color(black)("kJ roast")))))^(color(blue)("= 14% absorbed")) = "10,714.3 kJ"#

Now, the balanced thermochemical equation that describes this reaction looks like this

#"C"_ 3"H"_ (8(g)) + 5"O"_ (2(g)) -> 3"CO"_ (2(g)) + 4"H"_ 2"O"_ ((l))" "DeltaH_"rxn" = -"2217 kJ/mol"#

As you can see, the enthalpy change of reaction is equal to

#DeltaH_"rxn" = -"2217 kJ/mol"#

This tells you that for every #1# mole of propane that undergoes combustion, the reaction gives off--hence the minus sign--exactly #"2217 kJ"# of heat.

Moreover, you know that for every #1# mole of propane that undergoes combustion, the reaction produces #3# moles of carbon dioxide.

This means that when the reaction gives #"10.714.3 kJ"#, it also produces

#"10,714.3" color(red)(cancel(color(black)("kJ"))) * "3 moles CO"_2/(2217 color(red)(cancel(color(black)("kJ")))) = "14.5 moles CO"_2#

To convert this to grams, use the molar mass of carbon dioxide

#14.5 color(red)(cancel(color(black)("moles CO"_2))) * "44.01 g"/(1color(red)(cancel(color(black)("mole CO"_2)))) = color(darkgreen)(ul(color(black)("640 g")))#

The answer is rounded to two sig figs.