A 10.0 L flask contains 2.5 atm of ethane gas and 8.0 atm of oxygen gas at 28°C.The contents of the flask react until the limiting reactant is consumed.What is the pressure of carbon dioxide in the flask after the temperature returns to 28°C?
2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(g)

Question

Quizplus · Accepted Answer

The balanced chemical equation is $2C_2H_6(g) + 7O_2(g) \rightarrow 4CO_2(g) + 6H_2O(g)$. For every 2 moles of ethane, 7 moles of oxygen are required, and 4 moles of CO2 are produced. Ethane is the limiting reactant because it would require 7 moles of oxygen for every 2 moles of ethane, but there is more than enough oxygen present. Thus, all 2.5 atm of ethane react, producing $2.5 \times \frac{4}{2} = 5.0$ atm of CO2. However, the total pressure decreases due to the consumption of gases, so we must account for the pressure change. Initially, the total pressure is $2.5 + 8.0 = 10.5$ atm. After the reaction, 2.5 atm of ethane and 8.75 atm of oxygen (since $2.5 \times \frac{7}{2}$) are consumed, but this calculation does not directly affect the CO2 pressure calculation, which is based on the stoichiometry of the reaction. Given the stoichiometry, for every 2 moles of ethane, 4 moles of CO2 are produced, leading to a direct calculation of CO2 pressure as 5.0 atm, which seems to be a straightforward application of the stoichiometric ratios. However, the correct interpretation involves recognizing that the pressure of CO2 produced directly relates to the amount of ethane reacted, without needing to adjust for the total pressure change in the system post-reaction in this context. The mistake in the explanation is assuming additional steps that aren't required for determining the CO2 pressure based on the given reaction and initial conditions. The correct pressure of CO2 produced is indeed 5.0 atm, aligning with the stoichiometric outcome of the reaction based on the limiting reactant, ethane.

A 100 L Flask Contains 2