Asked by Anonymous
Calculate the partial pressure (in atm) of propane in a mixture that contains equal weights of propane (C3H8) and butane (C4H10) at 18°C and 746 mmHg
Answers
Answered by
DrBob222
P<sub>propane</sub> = X<sub>propane</sub>*P<sub>total</sub>
P<sub>total = 746/760 = ?atm (the problem asks for atm).
So all we need to do is to determine X<sub>propane</sub> and plug into the above.
molar mass C3H8 = about 44
molar mass C4H10 = about 58. YOu can do all of this more accurately.
X<sub>propane</sub> = (g/44)/[(g/44)+(g/58)]
Solve for X and substitute into the above. I get something like 420 mm (but the problem asks for atm).
P<sub>total = 746/760 = ?atm (the problem asks for atm).
So all we need to do is to determine X<sub>propane</sub> and plug into the above.
molar mass C3H8 = about 44
molar mass C4H10 = about 58. YOu can do all of this more accurately.
X<sub>propane</sub> = (g/44)/[(g/44)+(g/58)]
Solve for X and substitute into the above. I get something like 420 mm (but the problem asks for atm).
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.