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 3.39 x 102 L of methane and 1.70 x 102 L of oxygen?
This is a limiting reagent (LR) problem.
Determine the limiting reagent.
Use the coefficients in the balanced equation to convert mols of the LR to mols of the product.
Then convert mols of the product to grams by = mols x molar mass.
To find the maximum volume of carbon monoxide (CO) obtained, we need to calculate the limiting reactant first. The limiting reactant is the one that is completely consumed and determines the maximum amount of product that can be formed.
First, let's calculate the number of moles of methane (CH4) and oxygen (O2) given their volumes and assume an ideal gas behavior. We can use the ideal gas equation:
PV = nRT
Where:
P = pressure (constant)
V = volume
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (constant)
For methane (CH4):
V(CH4) = 3.39 x 10^2 L
R = 0.0821 L·atm/(mol·K) (at standard temperature and pressure, 0°C and 1 atm, so T = 273 K)
Assuming ideal gas behavior, we can rearrange the ideal gas equation to solve for n(CH4):
n(CH4) = PV(CH4) / RT
n(CH4) = (1 atm) (3.39 x 10^2 L) / (0.0821 L·atm/(mol·K) * 273 K)
n(CH4) ≈ 12.43 mol
For oxygen (O2):
V(O2) = 1.70 x 10^2 L
R = 0.0821 L·atm/(mol·K)
Assuming ideal gas behavior, we can rearrange the ideal gas equation to solve for n(O2):
n(O2) = PV(O2) / RT
n(O2) = (1 atm) (1.70 x 10^2 L) / (0.0821 L·atm/(mol·K) * 273 K)
n(O2) ≈ 8.82 mol
Now, let's determine the mole ratio between methane and carbon monoxide in the reaction:
2 CH4(g) + 3 O2(g) → 2 CO(g) + 4 H2O(g)
From the balanced equation, we can see that 2 moles of methane produce 2 moles of carbon monoxide. Therefore, the mole ratio of methane to carbon monoxide is 2:2, which simplifies to 1:1.
Since the mole ratio tells us that 1 mole of methane produces 1 mole of carbon monoxide, and we have 12.43 moles of methane available, we can say that we can produce a maximum of 12.43 moles of carbon monoxide.
Finally, let's calculate the maximum volume of carbon monoxide using the ideal gas equation:
n(CO) = 12.43 mol
R = 0.0821 L·atm/(mol·K) (at standard temperature and pressure, 0°C and 1 atm, so T = 273 K)
To find the volume, we rearrange the ideal gas equation:
V(CO) = n(CO)RT / P
Assuming the pressure is 1 atm:
V(CO) = (12.43 mol) (0.0821 L·atm/(mol·K)) (273 K) / (1 atm)
V(CO) ≈ 275.33 L
Therefore, the maximum volume of carbon monoxide that can be obtained from 3.39 x 10^2 L of methane and 1.70 x 10^2 L of oxygen is approximately 275.33 L.
To determine the maximum volume of carbon monoxide (CO) that can be obtained from the given amount of methane (CH4) and oxygen (O2), we need to use the stoichiometry of the reaction.
The balanced equation for the partial combustion of methane states that:
2CH4(g) + 3O2(g) > 2CO(g) + 4H2O(g)
From the equation, we can see that 2 moles of CH4 react with 3 moles of O2 to produce 2 moles of CO.
To solve the problem, we need to follow these steps:
Step 1: Convert the given volumes of methane and oxygen to moles.
To do this, we will use the ideal gas law equation:
PV = nRT
Given:
P = constant pressure (given as constant)
V = volume of gas (given in the question)
n = number of moles of gas (what we need to find)
R = ideal gas constant (constant)
T = constant temperature (given as constant)
Step 2: Calculate the number of moles for methane.
Using the ideal gas law equation, we know that the volume of methane (V) in liters is equal to the number of moles (n) multiplied by the ideal gas constant (R) multiplied by the constant temperature (T) in Kelvin:
n(CH4) = V(CH4) / (R * T)
Step 3: Calculate the number of moles for oxygen.
Similarly, we calculate the number of moles for oxygen using the same ideal gas law equation:
n(O2) = V(O2) / (R * T)
Step 4: Determine the limiting reactant.
The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product formed. To find the limiting reactant, compare the mole ratios of CH4 to O2. In this case, we compare:
2 moles of CH4 : 3 moles of O2
The limiting reactant will be the one with the smaller mole ratio.
Step 5: Calculate the moles of CO produced.
Since the limiting reactant is methane (CH4), we use the mole ratio from the balanced equation to determine the moles of CO produced:
n(CO) = n(CH4) * (2 moles of CO / 2 moles of CH4)
Step 6: Convert moles of CO produced to volume.
Using the ideal gas law, we can convert the moles of CO produced to the corresponding volume:
V(CO) = n(CO) * (R * T) / P
Now that you understand the step-by-step process, you can plug in the given values and calculate the maximum volume of carbon monoxide that can be obtained.