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
-------------------
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.
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
1 answer