The following equation represents the partial combustion of methane,CH₄. 2CH₄(g) + 3O₂(g) → 2CO(g) + 4H₂O(g)
At constant temperature and pressure,what is the maximum volume of carbon monoxide that can be obtained from 3.86 × 10² L of methane and 1.93 × 10² L of oxygen?

Question

Quizplus · Accepted Answer

To solve this problem, we need to use the concept of limiting reactants. We have 3.86 × 10² L of methane and 1.93 × 10² L of oxygen. Let's first convert these volumes to moles using the ideal gas law: nCH4 = PV/RT = (1 atm)(3.86 × 10²/22.4 L/mol)/(0.0821 L atm/mol K)(298 K) = 4.73 moles nO2 = PV/RT = (1 atm)(1.93 × 10²/22.4 L/mol)/(0.0821 L atm/mol K)(298 K) = 2.37 moles Now we need to determine which reactant is limiting. From the balanced equation, we see that every 2 moles of methane react with 3 moles of oxygen to produce 2 moles of carbon monoxide. Therefore, the ratio of moles of methane to moles of oxygen required is 2:3. If we have only 4.73 moles of methane, we would need 4.73 × (3/2) = 7.10 moles of oxygen to fully react with it. However, we only have 2.37 moles of oxygen, which is the limiting reactant. Using this information, we can calculate the maximum volume of carbon monoxide produced using the mole ratio from the balanced equation: nCO = nO2 × (2/3) = 1.58 moles VCO = nCO × RT/P = (1.58 mol)(0.0821 L atm/mol K)(298 K)/(1 atm) = 0.0389 m3 = 1.29 × 10² L Therefore, the correct answer is D, 1.29 × 10² L.

The Following Equation Represents the Partial Combustion of Methane,CH4

The Following Equation Represents the Partial Combustion of Methane,CH₄