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9 April, 10:09

Combustion of hydrocarbons such as propane (C3H8) produces carbon dioxide, a "greenhouse gas." Greenhouse gases in the Earth's atmosphere can trap the Sun's heat, raising the average temperature of the Earth. For this reason there has been a great deal of international discussion about whether to regulate the production of carbon dioxide. 1. Write a balanced chemical equation, including physical state symbols, for the combustion of gaseous propane into gaseous carbon dioxide and gaseous water. 2. Suppose 0.150kg of propane are burned in air at a pressure of exactly 1atm and a temperature of 12.0°C. Calculate the volume of carbon dioxide gas that is produced. Round your answer to 3 significant digits.

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  1. 9 April, 11:17
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    1. C₃H₈ (g) + 5O₂ (g) → 3CO₂ (g) + 4H₂O (g)

    2. The volume of CO₂ produced is 238 l.

    Explanation:

    Hi there!

    The chemical equation is the following:

    C₃H₈ (g) + 5O₂ (g) → 3CO₂ (g) + 4H₂O (g)

    The molar mass of propane is calculated as follows:

    mass of 1 mol carbon = 12 g

    mass of 1 mol hydrogen = 1 g

    mass of 1 mol propane 3 · (12 g) + 8 · (1 g) = 44 g

    If we have 150 g of propane, we will have (150 g · 1 mol / 44 g) 3.4 mol propane.

    Looking at the chemical equation, notice that 1 mol propane produces 3 mol CO₂. Then 3.4 mol of propane will produce:

    (3.4 mol C₃H₈ · 3 mol CO₂ / mol C₃H₈) 10.2 mol CO₂

    Using the Ideal Gas Law we can obtain the volume of CO₂ produced:

    P · V = n · R · T

    Where:

    P = pressure.

    V = volume

    n = number of moles.

    R = gas constant (0.082 atm · l / (K · mol))

    T = temperature in kelvin.

    V = n · R · T / P

    V = 10.2 mol · 0.082 atm · l / (K mol) · 285 K / 1 atm

    Notice that 12 °C = 273 + 12 = 285 K

    V = 238 l

    The volume of CO₂ produced is 238 l.

    Have a nice day!
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