How do you calculate the enthalpy of formation of butaine

 "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

yes it is balanced

b. th3 n4zc4 pl4t3 15 c0nv3rg1ng w1th the s0uth 4m3ric4n pl4t3.