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= 1211 kJ mol-1. Therefore the molar heat of combustion of ethanol is 1211 kJ mol-1 (The molar enthalpy of combustion of ethanol is 1211 kJ mol-1). (Note that the enthalpy change for the reaction is negative because the reaction is exothermic, so the enthalpy change for the reaction is -1211 kJ mol-1, ÎH = - 1211 kJ mol-1)
Confidence votes 13.1K. The density of ethanol is 0.789 g/mL. So 19.6 mL of ethanol weighs 0.789 g/mL * 19.6 mL = 15.464 g. The molar mass of ethanol is 2*12.011 + 6*1.008 + 15.999 = 46.069 g/mol. So, in 15.464 grams, there are 15.464 g / 46.069 g/mol = 0.33567 moles In one mole, there are 6.022*1023 molecules, so we have: 0.33567 moles * 6.022*1023 molecules/mole= 2.0214*1023 molecules. Using correct significant digits, that gives: 2.02*1023 molecules (or particles).