We can represent propane combustion by the following equation:
#C_3H_8(g) + O_2(g)rarr3CO_2(g) + 4H_2O(l) + 2220* kJ#
I took the heat of combustion values from this site. Your question does not even quote UNITS..........
And, yes, I know that we would normally write #DeltaH_"combustion"^@=-2220*kJ*mol.# We KNOW that hydrocarbon combustion is exothermic, so we can represent energy, #Delta#, as a reaction PRODUCT, for the purposes of answering this question.
And we have a #1000*L# of gas at #"STP"#, which of course we must treat as an ideal gas to get the molar quantity:
#n=(PV)/(RT)=(1*cancel"bar"xx10^3*cancelL)/(8.31xx10^-2*cancel"bar"*cancel(K^(-1))*mol^(-1)*cancelLxx273*cancelK)#
#=44*1/(mol)^-1=44*1/(1/(mol))=44*mol#.
We have #DeltaH_"combustion"^@=-2220*kJ*mol^-1#, and so we multiply the molar quantity by #DeltaH_"combustion"^@#
#=44*molxx-2220*kJ*mol^-1~=10^5*kJ#
Note that I have made assumptions with respect to #"standard conditions"#, which might be different to those specified by your syllabus; the UK syllabus is different to IB is different to the USA syllabus. I have ASSUMED, #"1 bar pressure"#, and #273*K#.
