How can I calculate heat of combustion and enthalpy using Hess's Law?

[Stephanus]

Dear PF Forum,
I have read this link in Wiki.
But I don't understand what it means. Can someone help me?

https://en.wikipedia.org/wiki/Heat_of_combustion
A: Heat of combustion of CH₄ is 50.09 MJ/kg
https://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_(data_table)
B: Enthalpy of water: -285.88 KJ/mol
C: Enthalpy of CO₂: -393.509 KJ/mol

Then, I'll try to do a simple calculation.

A: CH₄ heat of combustion is 50.09 MJ/kg.
What does it mean?
Does it mean that if we burn 1 kg of CH₄ completely
CH₄ + 2O₂ -> CO₂ + 2H₂O
It will produce 50.09 mega joules?

B: Enthalpy of water is 285.88 kj/mol
What does it mean?
If we combine 2 moles of Hydrogen and 1 moles of Oxygen, it will explode and gives 285.88 kilo joules?

C: Then I did some calculation...
The heat of combustion of
1 moles CH₄ + 2 moles O₂ -> 1 moles CO₂ + 2 moles H₂O
16 gr CH₄ + 32 gr O₂ -> 44 gr CO₂ + 36 gr H₂O will gives 50.09 MJ/kg * 16 gr = 801.44 KJ

The enthalpy of
1 moles H₂O: -285.88 KJ
1 moles CO²: -393.519 KJ

Combining those two:
2 moles H₂O: -571.76 KJ/mol
1 moles CO²: -393.519 KJ/mol
= 965.269 KJ

801.44 KJ ≠ 965.269 KJ

Where did I go wrong?
Or my understanding of the concept of heat combustion and enthalpy is wrong.
Thanks for any answer.

 

[Bystander]

Check the "phase" of water (liq/vap).

 

[Stephanus]

Check the "phase" of water (liq/vap).

Of course. Thanks. The numbers are close now.

 

[Chestermiller]

Of course. Thanks. The numbers are close now.

Liquid water vs water vapor accounts for only about half the difference. You also forgot to subtract the heat of formation of methane, which is -75 kJ/mole.

Chet

 

[James Pelezo]

You might also read about Hess's Law Equation. That is, if you haven't already done so. Heat of Rxn (or, Heat of Combustion in this case) = (Sum of Enthalpies of Formation of Products) - (Sum of Enthalpies of formation of Reactants) Google 'Enthalpy of Formation Tables' You'll get many 'energy of formation' values to apply.