The combustion of ethane:
2 C2H6 (g) + 7 O2 (g) ---> 4 CO2 (g) + 6 H2O (g)
At the same temperature and pressure, what is the maximum volume in liters of carbon dioxide that can be obtained from 15.9 L of ethane and 52.3 L of oxygen?
You can take a short cut when dealing with all gases. The short cut is that you can deal directly with volume and not convert to mols. But to complicate matters this is a limiting reagent problem. First convert L ethane to L CO2.
L CO2 = 15.9 L C2H6 x ( 4 mols CO2/2 mols C2H6) = ?
Then do the same with 52.3 L O2.
L CO2 = 52.3 L O2 x (4 mols CO2/7 mols O2) = ?
It is likely that the two values for volume CO2 will not agree; obviously one of them is wrong. In limiting reagent problems the smaller values is ALWAYS the correct one.
29.88 L
To find the maximum volume of carbon dioxide that can be obtained from the given amounts of ethane and oxygen, we need to first determine which reagent is limiting.
Step 1: Identify the limiting reagent
To determine the limiting reagent, we compare the moles of ethane to the moles of oxygen using their balanced stoichiometric coefficients.
Given:
15.9 L of ethane
52.3 L of oxygen
Since we need to compare moles, we need to convert the volumes to moles using the ideal gas law equation:
PV = nRT
Where:
P is the pressure (assumed constant)
V is the volume
n is the number of moles
R is the ideal gas constant
T is the temperature (assumed constant)
For simplicity, assuming constant pressure and temperature, let's use the ideal gas constant (R = 0.0821 L·atm/(mol·K))
Converting ethane volume to moles:
n_ethane = (V_ethane * P) / (R * T)
Converting oxygen volume to moles:
n_oxygen = (V_oxygen * P) / (R * T)
Since we have the same temperature and pressure for both gases, we can calculate the moles directly.
Given: T, P, R = constant
V_ethane = 15.9 L
V_oxygen = 52.3 L
n_ethane = V_ethane
n_oxygen = V_oxygen
Step 2: Calculate the moles of carbon dioxide produced
Now that we have the number of moles of ethane (n_ethane) and oxygen (n_oxygen), we can calculate the theoretical moles of carbon dioxide (CO2) produced based on the stoichiometric coefficients in the balanced equation:
2 C2H6 (g) + 7 O2 (g) ---> 4 CO2 (g) + 6 H2O (g)
From the balanced equation, we can see that for every 2 moles of ethane, we produce 4 moles of carbon dioxide.
So the moles of carbon dioxide produced (n_CO2) can be calculated as:
n_CO2 = (n_ethane / 2) * 4
Step 3: Calculate the volume of carbon dioxide
Finally, we can calculate the volume of carbon dioxide (V_CO2) based on the moles of carbon dioxide produced (n_CO2) and the ideal gas law:
V_CO2 = (n_CO2 * R * T) / P
Given:
R = 0.0821 L·atm/(mol·K) (ideal gas constant)
T = constant (assumed)
P = constant (assumed)
Substituting the values:
V_CO2 = (n_CO2 * R * T) / P
Step 4: Calculate the maximum volume of carbon dioxide
By substituting the calculated values of n_CO2, R, T, and P into the equation, we can find the maximum volume of carbon dioxide that can be obtained.
Therefore, the maximum volume in liters of carbon dioxide that can be obtained can be calculated using the given formula.
Please note that the temperature and pressure values need to be provided in order to determine the exact value.