Propane (C3H8) which is used as fuel in BBQ tanks often contains some butane(C4H10). A sample of a fuel mixture known to contain both propane and butane occupies a certain volume at a total pressure of 30 mmHg. The sample is then burned
completely to produce CO2 and H2O. The CO2 alone is collected and its pressure found to be 93 mmHg in the same volume and at the same temperature as the original
mixture. What was the mole fraction of propane in the original sample mixture before combustion?
C3H8 + 5 O2 3 CO2 + 4 H2O
C4H10 + 13/2 O2 4 CO2 + 5 H2O
A little long but not that complicated.
Use PV = nRT and solve for n = total number of mols C3H8 + C4H10. You have P and R but not T and V. I would use convenient numbers for those two and you can do that since the SAME numbers will be used in the CO2 part of the problem. Using V = 1L and T = 300K, I obtained 0.001603 but you should confirm that. Remember P must be in atm if R is 0.08206 L*atm/mol*K.
Let X = mols C3H8
and Y = mols C4H10
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Then equation 1 is X + Y = 0.001603
The CO2 part is done much the same way.
Use PV = nRT and solve for total mols CO2. I obtained 0.004971 but you should confirm that. Use the same T, R, and V you used in the first part. Here mols CO2 from C3H8 + mols CO2 from C4H10 = total mols CO2. Looking at the equation, we make equation 2 for this problem in terms of X and Y as follows:
3X(which = mols CO2 produced from C3H8) + 4Y(which = mols CO2 produced from C4H10) = total mols or
equation 2 is 3X + 4Y = 0.004971
Now solve equation 1 above and equation 2 above simultaneously for X and Y.
Then mols fraction C3H8 = (mols X/total mols) or mols (X/0.001603) = ?
I obatained something like 0.9 but that's a close estimate.
Post your work if you get stuck.
To determine the mole fraction of propane in the original sample mixture, we need to calculate the number of moles of propane and butane present in the sample before combustion.
Let's assume the volume of the sample is 1 liter.
Step 1: Calculate the mole fraction of CO2 in the final mixture.
Since the pressure of CO2 is given as 93 mmHg, which is the same as the total pressure of the original mixture (30 mmHg), we can conclude that all the CO2 produced in the combustion reaction remains in the sample container.
The mole fraction of CO2 (X_CO2) can be calculated using the ideal gas law:
PV = nRT
Where:
P = Pressure of the gas (93 mmHg)
V = Volume of the gas (1 L)
n = Number of moles of the gas
R = Ideal gas constant
T = Temperature (assumed constant)
Since the pressure and volume are constant, we can infer that the number of moles of CO2 is proportional to the mole fraction of CO2.
X_CO2 = n_CO2 / (n_CO2 + n_H2O)
Since the combustion reaction equation shows that 3 moles of CO2 are produced for every mole of propane burned, we can write:
X_CO2 = 3n_propane / (3n_propane + 4n_H2O)
But, since the volume and temperature are the same for both the original mixture and the combustion products, the number of moles of water (n_H2O) in both cases will be the same. Therefore, we can eliminate it from the equation:
X_CO2 = 3n_propane / (3n_propane + 4n_H2O) = 93/30
Step 2: Determine the number of moles of butane.
From the combustion reaction equation, we can see that 4 moles of CO2 are produced for every mole of butane burned:
4n_butane = n_CO2
Step 3: Calculate the number of moles of propane.
To determine the number of moles of propane, we can use the following relation between the mole fraction and the number of moles:
X_propane = n_propane / (n_propane + n_butane)
Since the mole fraction of propane can be represented as (1 - X_CO2), we can write:
1 - X_CO2 = n_propane / (n_propane + n_butane)
Substituting the value of X_CO2 from step 1:
1 - (93/30) = n_propane / (n_propane + n_butane)
30/30 - 93/30 = n_propane / (n_propane + n_butane)
(30 - 93)/30 = n_propane / (n_propane + n_butane)
-63/30 = n_propane / (n_propane + n_butane)
Step 4: Calculate the mole fraction of propane.
Since we want to find the mole fraction of propane (X_propane), we can rearrange the equation from step 3:
X_propane = n_propane / (n_propane + n_butane)
Rearranging further:
X_propane = (-63/30) / (1 - (-63/30))
Simplifying:
X_propane = (-63/30) / (30/30 + 63/30)
X_propane = (-63/30) / (93/30)
X_propane = -63/93
X_propane ≈ -0.68
Since mole fractions cannot be negative, we know that our calculations are incorrect. Please check the data or equations provided and ensure they are accurate.
To find the mole fraction of propane (C3H8) in the original sample mixture, we need to use the ideal gas law and the stoichiometry of the combustion reaction.
1. Determine the moles of CO2 produced:
From the balanced equation, we can see that 1 mole of C3H8 produces 3 moles of CO2. Since the pressure of the CO2 collected is 93 mmHg, we can convert it to atm by dividing by 760 mmHg and use the ideal gas law to find the moles:
n_CO2 = (P_CO2 * V) / (R * T)
where P_CO2 is the pressure of CO2, V is the volume of the system, R is the ideal gas constant, and T is the temperature in Kelvin.
2. Determine the moles of butane (C4H10) in the sample before combustion:
From the balanced equation, we can see that 1 mole of C4H10 produces 4 moles of CO2. Using the same process as above, find the moles of CO2 produced by the combustion of butane.
3. Determine the moles of propane (C3H8) in the sample before combustion:
Since we know the total pressure before combustion is 30 mmHg, we can assume that the remaining pressure (30 mmHg - 93 mmHg) is due to the unreacted propane. Convert this pressure to atm and find the moles of propane using the ideal gas law.
4. Calculate the mole fraction of propane:
Mole fraction = Moles of propane / Total moles of fuel
where Total moles of fuel = Moles of propane + Moles of butane
Using this information, you can calculate the mole fraction of propane in the original sample mixture before combustion.