What was the final total pressure in the cylinder (in atm) after the reaction, while it was still at 125°C?

Question

A 11.3 mL sample of liquid ethanol (C₂H₅OH, density = 0.789 g/mL) was injected into a 10.0 L cylinder containing O₂ at a pressure of 4.50 atm and a temperature of 25°C. The cylinder was heated to 125°C and a spark was used to ignite the ethanol, which was completely combusted. What was the final total pressure in the cylinder (in atm) after the reaction, while it was still at 125°C?

Answer 1

Let's first look at what happens when you combust 11.3 ml liquid C₂H₅OH.

C₂H₅OH(l) + 3O₂(g) ==> 2CO₂(g) + 3H₂O(g/l) ... balanced equation. Note that you generate 2 moles CO₂ gas and depending on the final pressure, the H₂O can be a gas or a liquid. I'll assume it to be a liquid and not include it in the calculations assuming the pressure is high enough for it to condense into liquid.

moles C₂H₅OH = 11.3 ml x 0.789 g/ml x 1 mol/46 g = 0.1938 moles C₂H₅OH

moles O₂ present = n = PV/RT = (4.5)(10)/(0.0821)(298) = 1.84 moles O₂

O₂ is in excess.

moles CO₂ produced = 0.1938 mol C₂H₅OH x 2 mol CO₂/mol C₂H₅OH = 0.3876 mol CO₂

O₂ left over after combustion: 0.194 mol C₂H₅OH x 3 mol O₂/mol C₂H₅OH = 0.582 mol O₂ used

1.84 - 0.582 = 1.26 mol O₂ left over

Total moles of gas after combustion = 1.26 mol O₂ + 0.39 mol CO₂ = 1.65 moles of gas

PV = nRT

P = nRT/V = (1.65)(0.0821)(398)/10

P = 5.39 atm total pressure