[WhoCares357]

Use a combustion reaction of liquid ethanol in air as a source of work, assuming T=25 degrees Celsius, air pressure 1 atm and all products are gases with partial pressure of 10 atm each. How high can one lift a 1000kg weight by burning 100g of ethanol. (Oxygen is 21% by volume of air.)

I started by writing the equation:
C2H5OH(l)+3O2(g) -> 3CO2(g) + 3H2O(g)
100g of ethanol is 2.17 moles of ethanol; therefore there are 2.17 moles of ethanol, 6.51 (2.17*3) moles of O2, CO2, and H2O

25 degrees Celsius is 278.15 Kelvin

Work = dG = dH - TdS

Since the question states this is at standard pressure and temperature, I looked up the delta H and delta S for each product and reactant. I subtracted the reactants from the products and got the following answers:
dH = -3520.391kJ, dS = .935834kJ/K, T = 278.15K

I plugged them into the Work equation and got Work = -3799.41 kJ.

This is where I am stuck. I don't know how to use the information I have to find the distance. I know that Work = Force * Distance = -Pressure * change in Volume. I also know that Force = Mass * Acceleration. However, I do not know what the Force would equate to (how would I find the acceleration?). Am I right in saying that the Pressure is 20atm and the change in Volume is equal to 6.51moles of CO2 and H2O converted into Liters?

Where do I go from here?

Thank you for any help.