The heat of combustion of propane is -2,220.1 kJ/mol

Question

calculate the heat of formation, delta Hf, of propane given that delta Hf of H20= -285.3 kJ/mol and delta Hf of CO2 = -393.5 kJ/mol

Answer 1

C3H8 + 5O2 ==> 3CO2 + 4H2O
-2220.1 kJ/mol = (n*sum delta Hfproducts) - (n*sum delta Hfreactants).
The only unknown is delta Hf C3H8. You have the values for H2O and CO2. O2, of course, is zero.

Answer 2

so we multiply the sum by the total number of moles, or by moles of reactant and product separately?

Answer 3

The products are CO2 and H2O.
So it's [(3*deltaH CO2) + (4*deltaH H2O)] - (1*deltaH C3H8) = -2220.1 kJ and solve for deltaH propane.