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

Hi Marleni, since you want to solve for the final pressure in the container, you'll want to use the ideal gas law (PV = nRT ). Since the final temperature is provided and the volume of the container doesn't change, all you need to find the final pressure is the final moles of total gas within the container.

Here's a breakdown on how to get there:

1.) balance the chemical reaction for ethanol's combustion (_C₂H₅OH + _O₂ -> _CO₂ + _H₂O)

2.) calculate the starting quantity of moles of ethanol. To do this, multiply the density of ethanol by the volume to find the mass (d x v = m), then divide the mass of ethanol present by ethanol's molar mass to arrive at the number of moles (m/M = n).

3.) The number of moles of ethanol is equal to the coefficient in front of C₂H₅OH in your balanced reaction formula. Knowing the number of moles that corresponds to that coefficient you can solve for the number of moles that correspond to the other coefficients in your balanced equation. Use that information to calculate the moles of O₂ consumed in the reaction and moles of CO₂ produced.

4.) Calculate the moles of O₂ remaining after the reaction by using n = (PV)/(RT) to find the original number of moles of O₂. Then subtract from that the number of moles of O₂ consumed in the reaction calculated in the previous step. Then add the moles of CO₂ produced to the remaining moles of O₂.

5.) Plug the total moles of gas after combustion into (nRT/V) = P to arrive at the final pressure.

Hope this helps!

Answer 2

P1 = 4.5 atm

V1=V2

T1=25C = 298 K

T2=125 C = 398 K

   C2​H5​OH+3O2​→2CO2​+3H2​O

moles   1     3      2    3

mass   46    96     88   54

 1) find moles of ethanol

Density of ethanol = 0.789 g/ml

11.3ml×(0.789g/ml​)/(46.07g/mole​)=0.194 moles(C2​H5​OH)

2) find initial moles of O2. Use Ideal Gas Law

PV=nRT

n=PV/RT​=( 4.50atm×10.0L​)/(0.082 L-atm/(mol K)​×(273+25)K)=1.84 moles

3) Find limiting reagent

0.194 moles(C2​H5​OH)​ < 1.84/3 moles(O2​)​

C2​H5​OH - limiting reactant

4) Find moles of O2 after reaction

The moles of O2 consumed in the reaction:

0.194moles(C2​H5​OH)×3moles(O2​)​/1mole(C2​H5​OH)=0.581 moles(O2​)

The moles of O2 left after the reaction was completed:

1.84 moles(O2​)−0.58 moles(O2​)=1.26moles(O2​)

5) Find moles of CO2

0.194moles(C2​H5​OH)×2moles(CO2​)​)/(1mole(C2​H5​OH)=0.39moles(CO2​)

6) Find moles of H2O

0.194moles(C2​H5​OH)×3moles(H2​O)/(1mole(C2​H5​OH)​)=0.58moles(H2​O)

7) Find the total number of moles of gases:

1.26+0.39+0.58=2.23 moles

8) Find total pressure:

PV=nRT

P=nRT/V​= 2.162moles×0.082 L×atm/(mol×K)​×(273+125)​/10.0L=7.28 atm

If all H2O is in the form of liquid,

9) Find the total number of moles of gases:

1.26+0.39=1.65 moles 1.65

10) Find total pressure:

PV=nRT

assuming the volume occupied by water is negligible

Water volume 10.44 mL

V= 9.98956

V =~ 10 L

P=nRT/V​= 2.162moles×0.082 L×atm/(mol×K)​×(273+125)​/10.0L=5.38 atm

5.38 atm

545.5 kPa

Would some of the water evaporate at this pressure???

Vapor pressure at 125 C 240 kPa

2.37 atm

If pressure is above 2.4 atm, H2O will be in the liquid form

Therefore, all H2O will be in liquid form!

Pressure in the cylinder 5.38 atm