[Frederick95]

Question: Use tabulated standard enthalpies of formation to calculate a theoretical value for the molar enthalpy of combustion of acetone.


THE FOLLOWING IS MY WORK PLEASE TELL ME IF THERE ARE ANY MISTAKES. THANK YOU

1.   Write a balanced equation: C3H6O(l) + 4O2(g) --> 3CO2(g) + 3H2O(l)

2.   Under each compound, write out the enthalpy of formation, ignoring the number of moles.
                      C3H6O(l) + 4O2(g) --> 3CO2(g) + 3H2O(l)
                      -248.1       0.0           -393.5       -285.8

3.   Put the enthalpy values into the equation but now you have to take the number of moles into account:

        ∆H = ∑n∆H°f(prod.) - ∑n∆H°f(react.)
 
                     = [3(-393.5) + 3(-285.] – [(-248.1) + 4(0.0)] kJ/mol
                     = [-2037.9 – (-248.1)] kJ/mol
                     = -1789.8 kJ/mol

[ARGOS++]

Dear Frederick95;

I have not checked  where you got the other values from, but otherwise it looks ok for me.

Good Luck!
                    ARGOS⁺⁺