If 5.00 L of propane is burned in 21.0 L of oxygen,what volume of carbon dioxide is produced? Assume that the temperature of the reactants and products is 25°C and the pressure of the system remains constant at 1.0 atm. C₃H₈(g)+ 5O₂(g)→ 3CO₂(g)+ 4H₂O(l)

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Quizplus · Accepted Answer

The balanced chemical equation for the combustion of propane (C3H8) with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O) is: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l). According to the stoichiometry of the reaction, 1 mole of propane reacts with 5 moles of oxygen to produce 3 moles of carbon dioxide. Given that 5.00 L of propane is burned in 21.0 L of oxygen, the limiting reactant is propane because it requires 25 L of oxygen to completely react (5 times the propane volume), but only 21.0 L is available. Since the reaction is under constant temperature and pressure, the volume ratio of gases can be directly related to their mole ratio. Therefore, burning 5.00 L of propane will produce 3 times the volume in carbon dioxide, which is 15.0 L. However, since oxygen is in excess (only needing 5 L for every 1 L of propane), the actual limiting factor is the amount of propane available. Thus, the correct calculation based on the stoichiometry and the given conditions should conclude that 15.0 L of CO2 is produced, indicating a mistake in the explanation. The correct volume of CO2 produced should directly relate to the stoichiometry of the reaction, which indicates 3 volumes of CO2 for every 1 volume of propane when oxygen is in excess.

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