Question

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?

Answer 1

Answer:

             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,

Data:

       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