Please help me with chemistry math problems

Question

What is the heat required to vaporize 90.8 g of liquid ethanol, C₂H₅OH, at its boiling point? The ΔH° of vaporization of ethanol is 39.3 kJ/mol. Use a molar mass with at least as many significant figures as the data given.

Answer in kJ

Answer 1

The given enthalpy of vaporization of 39.3 kJ/mol indicates that vaporization of one mole of ethanol requires 39.3 kJ of heat. We can use the molar mass of ethanol to convert the given mass to moles and then use this 39.3 kJ/1 mol relationship as a conversion factor to relate the number of moles of ethanol present to the quantity of heat required to vaporize it.

C₂H₅OH: 2(12.01) + 5(1.01) + 16.0 + 1.01 = 46.08 g/mol

90.8 g C₂H₅OH x (1 mol C₂H₅OH/46.08 g C₂H₅OH) = 1.97... mol C₂H₅OH

1.97... mol C₂H₅OH x (39.3 kJ/1 mol C₂H₅OH) = 77.4 kJ

77.4 kilojoules of heat are required to vaporize this sample of ethanol.