Using enthalpies of formation, calculate the quantity of heat produced when 14.0 g of propane is completely combusted in air under standard conditions. assume that liquid water is forming

The balanced chemical reaction here is:

C3H8(g) + 5O2(g) --> 3CO2(g) + 4H2O(l)

Calculating for the molar change in enthalpy:

Enthalpy change = 3(-393.5) + 4(-285.83) - (-103.85) = -2220.0 kJ/mole 

Now, propane has a molar mass of 44g/mol. So, for us to calculate the heat released by burning 14 grams of propane, we have: 
-2220 kJ/mole * ( 14g / 44g/mol) = -706.36 kJ 

-706.36 kJ 

(negative means heat is released)