The combustion of propane occurs when propane interacts with oxygen gas to produce water and carbon dioxide in the following reaction:

C3H8 (g) + 5O2 (g) = > 3CO2 (g) + 4H2O (l)

If 2 moles of propane C3H8 (g) react at 373 K in a volume of 2.5 L, what is the resulting pressure in the cylinder from the CO (g) produced in the reaction?

Pressure = 73.49 atm

Explanation:

The balance chemical equation is as follow,

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

According to balanced equation,

1 mole of C₃H₈ on combustion gives = 3 moles of CO₂

So,

2 moles of C₃H₈ on combustion will give = X moles of CO₂

Solving for X,

X = 2 moles of C₃H₈ * 3 moles of CO₂ : 1 mole of C₃H₈

X = 6 moles of CO₂

Now, in second step we will calculate the the pressure exerted by CO₂ at 2.5 L volume and 373 K temperature. For this we will use Ideal gas equation assuming the gas is acting as an ideal gas. Therefore,

dа ta:

Temperature = T = 373 K

Volume = V = 2.5 L

Moles = n = 6 mol ∴ As calculated above.

Gas Constant = R = 0.0821 atm. L. mol⁻¹. K⁻¹

Formula Used:

P V = n R T

Solving for P,

P = n R T / V

Putting Values,

P = 6 mol * 0.0821 atm. L. mol⁻¹. K⁻¹ * 373 K : 2.5 L

P = 73.49 atm