What amount of heat, in kJ, is required to vaporize 115.00 g of ethanol ...

Question

What amount of heat, in kJ, is required to vaporize 115.00 g of ethanol (C₂H₅OH)? (∆Hvap = 43.3 kJ/mol)

Answer 1

Hello, Ina,

The units of the heat of vaporization tell us how to solve for the heat required. 43.3 kJ/mol

It requires 43.3 kiloJoules (kJ) to vaporize one mole of ethanol. There is no temperature change in this question, just a phase change to go from liquid to gas.

If we know the moles of ethanol that are vaporized, we can find the energy required.

We need the molar mass of ethanol. I find 46.07 g/mole. Find moles by dividing the mass by the molar mass.

(115.00 g C₂H₅OH)/(46.07 g C₂H₅OH/mole C₂H₅OH) = 17.35 kJ

Bob

Answer 2

You can simply use dimensional analysis, after converting (a) ∆Hvap to kJ/g or (b) converting 115.00 g to moles.

a). molar mass ethanol = 46.07 g/mol

∆Hvap = 43.3 kJ/mol x 1 mol/46.07 g = 0.940 kJ/g

115.00 g x 0.940 kJ/g = 108 J (3 sig. figs.)

b) 115 g ethanol x 1 mol/46.07 g = 2.496 moles

2.496 moles x 43.3 kJ/mol = 108 J (3 sig. figs.)