Enthalpy of Formation: Solve CO + O_2 to CO_2

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To solve the enthalpy of formation problem for the combustion of carbon monoxide (CO) to carbon dioxide (CO2), the change in enthalpy (ΔH) must be calculated using the provided molar enthalpies. The reaction is 2CO(g) + O2(g) → 2CO2(g), with CO having an enthalpy of -110.5 kJ/mol and CO2 at -393.5 kJ/mol. The correct formula for ΔH is ΔH = [2(-393.5 kJ)] - [2(-110.5 kJ)], which simplifies to -673 kJ for the reaction. Multiplying this value by the number of moles (13.39) will yield the total heat evolved during the combustion process. Proper calculation should confirm the correct energy output in kJ.
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Hi, I have a question involving enthalphy of formation. I tried doing this problem but I couldn't seem to get the right answer. I was wondering if anyone here may please help me out. Thanks.

Question: How much heat is evolved when 13.39 moles of CO(g) is burned in excess oxygen.
2CO(g) + O_2(g)\rightarrow 2CO_2(g)
Molar enthalpies are
CO=-110.5 kJ/mol
O_2=0.0
CO_2=-393.5 kJ/mol
 
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you figure out the Change of Enthalpy (change of H) of overall equation, and then multiply by 13.39 moles of CO to get energy (KJ).

just to make sure...

Change of H = [(2)(-110.5KJ)] -[(2)(-110.5KJ)] = ?? KJ

then mult ?? KJ by 13.39 mol

does it give you the right answer?? would like to know
 
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