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24 April, 18:21

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?

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  1. 24 April, 18:39
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    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
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