For NH3, 0.81 g/cc x 1000 cc = 810 grams.
810g/17 = ? mols.
2NH3 + 3/2 O2 ==> N2 + 3H2O
dHrxn = (3*dHf H2O + 0) - (2*dH NH3) = ? kJ/mol
You need to add in 23.2 to that to go from liquid NH3 to burned NH3.
Then total dHrxn (including vaporization) x mols NH3 = kJ.
Do the same thing for CH3OH but I see no number for correction from liquid to gas.
Ammonia (NH3) boils at -33oC; at this temperature it has a density of 0.81 g/cm3. The enthalpy of formation of NH3(g) is -46.2kj/mol, and the enthalpy of vaporization of NH3(l) is 23.2 kJ/mol. Calculate the enthalpy change when 1 L of liquid NH3 is burned in air to give N2(g) and H2O(g). How does this compare with ΔH for the complete combustion of 1 L of liquid methanol, CH3OH(l)? For CH3OH(l), the density at 25oC is 0.792 g/cm3, and ΔHof equals -239 kJ/mol.
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