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I was looking in the solutions manual for problem 57 and I was curious to know how they got the equations for C2H2, C2H6, and H2? Were we just supposed to know them? These equations were used to calculate enthalpy using Hess's Law.
These are standard enthalpies of combustion. For combustions reactions, it's always in the same form: some hydrogen/hydrocarbon burned in excess O2 gas yields to CO2 gas and H2O liquid. Keep in mind that these enthalpy values are based on one mole of C2H2, C2H6, and H2. So when you do manipulate these equations to cancel out, if you have a coefficient that's not one, you need to multiply that coefficient to the standard enthalpy of formation given for that compound. I hope this makes sense.
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