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 6.62 × 10² L of methane and 3.31 × 10² L of oxygen?

Question

Quizplus · Accepted Answer

According to the balanced equation, 2 moles of methane react with 3 moles of oxygen gas to form 2 moles of carbon monoxide and 4 moles of water vapor. We need to find the maximum volume of carbon monoxide, which is directly proportional to the number of moles of carbon monoxide formed.

From the given information, we have:
Volume of methane = 6.62 × 10^12 L
Volume of oxygen = 3.31 × 10^12 L

Using the ideal gas law, we can convert the volumes of methane and oxygen to the number of moles of each gas:
n(CH4) = (PV)/(RT) = [(1 atm)(6.62 × 10^12 L)]/[(0.0821 L·atm/mol·K)(298 K)] = 2.43 × 10^11 mol
n(O2) = (PV)/(RT) = [(1 atm)(3.31 × 10^12 L)]/[(0.0821 L·atm/mol·K)(298 K)] = 1.22 × 10^11 mol

According to the balanced equation, the ratio of the number of moles of methane to carbon monoxide is 2:2 or 1:1. So, the maximum number of moles of carbon monoxide that can be formed is also 2.43 × 10^11 mol.

Using the ideal gas law again, we can convert the number of moles of carbon monoxide to its volume:
V(CO) = n(CO)(RT)/P = (2)(0.0821 L·atm/mol·K)(298 K)/(1 atm) = 0.49 L

Therefore, the maximum volume of carbon monoxide that can be obtained is 0.49 L or 2.21 × 10^12 L, which is closest to choice D.

The Following Equation Represents the Partial Combustion of Methane,CH4

The Following Equation Represents the Partial Combustion of Methane,CH₄