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,

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

< > according to newton's third law of motion, forces always act in equal but opposite pairs. another way of saying this is for every action, there is an equal but opposite reaction.< >