[Amynare]

Hi can some one please help me with the following calculations of the heats of combustion for the following fuels, I'm a little rusty.


Magnesium Mg The equation for the burning of magnesium is: Mg + O2 → 2MgO


Carbon monoxide CO The equation for the burning of CO is: 2CO + O2 → 2CO2

[billnotgatez]

I rewrote your formulas with proper formatting and a correction

 2Mg + O₂ → 2MgO


 2CO + O₂ → 2CO₂

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[Amynare]

Have attempted working out how to calculate the heats of combustion for the following fuels. Is this correct? Unable to work out how to to make the atom numbers smaller, see post below. Any help would be most appreciated, thanks.

2Mg + O2 → 2MgO

(2x150) + 0  /     (2x150) + 0
 300kJ           /     300kJ

Difference of 0kJ



2CO + O2 → 2CO2 

(2x717) + 249 + 0      /       (2x717) + 0
 1683kJ                        /      1437kJ

Difference of 246kJ

(2x12) + 16 = 40

Thus burning one mole of Magnesium (40g) would release 246kJ of heat energy

[mjc123]

This is nonsense. The numbers you are using appear to be heats of formation of gaseous atoms. But this means formation of the atoms from the elements - not from compounds. E.g. 717 kJ/mol is the heat of forming C atoms from solid carbon - not from CO or CO₂. So you can't use these numbers in the way you are trying to. And heats of atomisation of compounds are not generally known or tabulated, as far as I know, so this isn't a good way to tackle the problem.
What is tabulated for many compounds is the heat of formation, and you can use this to work out your heats of combustion, e.g.
2Mg + O₂  2MgO   ΔH = 2ΔH_f(MgO) - 2ΔH_f(Mg) - ΔH_f(O₂). And remember that the heat of formation of elements is zero.
The above equation will give you the heat of combustion of 2 mol Mg, so if you want it per mol of Mg, divide by 2.