Chemistry

Energy, q, w, Enthalpy, and StoichiometrySource(s):
Consider the combustion of liquid methanol, CH3OH (l).

CH3OH (l) +3/2O2 (g) = CO2 (g) + 2H2O(l) H = -726.5 kJ
Part A:
What is the enthalpy change for the reverse reaction?
Answer I got right: 726.5 kJ
Part B:
Balance the forward reaction with whole-number coefficients:
CH3OH (l) +3/2O2 (g) = CO2 (g) + 2H2O(l) H
Part C:
What is for the reaction represented by this equation?
Answer : I am stuck. ?????

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1. Opps I left off what is H
Part C:
What is H for the reaction represented by this equation?
Answer : I am stuck. ?????

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2. I think it is this but not sure.
+ 726.4 kJ

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3. I'm stuck but not as much as before. I keep looking for WORDS like combustion, equilibrium, etc but the "ans to 4 s.f." means they want a number. Perhaps the number is -726.5kJ/mol*2 = -1453 kJ. I'm still in the dark if this isn't right.

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4. Now it makes sense with the H. The delta H for the doubled equation is 2*delta H for the original and the -1453 kJ should give you the correct answer.

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5. thank you that is the answer. -1453

It was driving crazy.

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6. I did not * 2 just one simple step will mess ya up lol.. thanks for your help... your awesome.

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8. Finally achieved a gr8 success my ans is -239

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