How do you calculate the enthalpy of formation of butaine

Answer: "Calculate the enthalpy of formation of butane, C4H10, using the balanced chemical equation and tables displaying information about enthalpy of formation/combustion. Write out the solution according to Hess's law." That's all I'm given. The key says that the answer is - 125.4 kJ. C4H10 + 6.5 O2 = 4CO2 + 5 H20 The heat of combustion for one mole of butane is - 2877 K. Heats of formation from table C02 = - 393.5 Kj/mole, H20 (g) - 241.8 Kj/mole. Heat of combustion of butane = sum heats of formation products minus heats of formation reactants - 2877 Kj = 4 (-393.5Kj) + 5 (-241.8Kj) minus heat or enthalpy of formation butane.

The enthalpy formation oxygen an element is assigned a value of zero.

-2877 Kj = - 1582 Kj + - 1209 Kj minus X

-2877 = - 2791 Kj minus X

X = - 86 Kj. The heats of formation for various compounds vary a little from table to table, so this does not quite agree with your answer.

This is just an example