Ethanol burns to produce 1300.0 kJ/mol and has a heat capacity of 65.6 J/mol·K. If Dr. Porter has 1.00 L of liquid ethanol (density = 0.789 g/mL), what is the minimum mass of fuel that should be removed and burned to heat the remaining ethanol from 25 °C to 90 °C? (Assume the equilibrium occurs in the gas phase.) Hint: Set-up an equation with the unknown X = ‘mass of ethanol’, where when units are canceled, you’re left with Joules = Joules.

Answer 1

I would do this although it may be the long way around. I would convert all of the "moles" to grams.
1300 kJ/mol = 1300000/46.069 = ??J/g for the heat of combustion.
65.6 J/mol = 65.6/46.00 = xx J/g for the heat capacity (specific heat) of ethanol.
1.00 L ethanol = 1000 mL and that has a mass of 1000 x 0.789 g/mL = xx grams.

Let X = mass ethanol removed.
Xg*heat combustion = (mass ethanol-X)*specific heat ethanol* (Tfinal-Tinitial).
Solve for X and 789-X.
I obtain something a little less than 3 grams to be removed and that will heat the remainder from 25 to 90.
Post your work if you get stuck.

Answer 2

Again Melissa?