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Topic: Molar flow rate and mole fraction  (Read 29197 times)

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Offline dillonv

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Molar flow rate and mole fraction
« on: February 16, 2009, 04:20:03 PM »
I'm having some trouble solving this problem. I was wondering if someone could give me some guidance?

"A gas containing equal parts of methane, ethane, and ammonia flows at a constant rate through a laboratory water-based absorption unit, which absorbs 96% of the ammonia and retains it in the liquid. No methane or ethane is absorbed into the water and no water evaporates into the gas. Initially, there was exactly 5.00 kg of water in the absorber, and at the end of 4 hours of operation, the liquid mass is 5.25 kg. Calculate the molar flow rate (mol/hr) of the gas stream coming into the absorber, and the mole fraction of ammonia in the exit gas stream."

Here is what I have tried:
To calculate the molar flow rate, I subtracted 5.00kg from 5.25kg, which gives .25 kg. I assumed this .25kg to be ammonia, based on the information in the problem. I then divided .25 by 4 hours, which gives .0625 kg/hr. I then converted .0625 kg ammonia to moles of ammonia, but I did not get the correct answer. The answer for the first part is supposed to be 11.48 moles/hr.
I'm not even sure how to start answering the second part of the question.

Any help would be much appreciated.

Offline ARGOS++

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Re: Molar flow rate and mole fraction
« Reply #1 on: February 17, 2009, 08:13:03 PM »

Dear dillnov;

Your start is ok!; - expect that you have only 96% of NH3 absorbed. How much is 100%?
Then you translate "100%" into moles NH3 per hour, and finally you multiply this value by the factor 3.0 to get the total molar flow rate.  Do you know where this 3.0 are caused by?

I hope to have been of help to you to start.
Good Luck!
                    ARGOS++

Offline oxygenfluid

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Re: Molar flow rate and mole fraction
« Reply #2 on: February 17, 2009, 08:59:16 PM »
Hello dillonv,

As ARGOS++ said only 96% of ammonia is absorved by the water, so you only calculated the flow that was absorved.

For the second part of the question;
molar fraction -> x=n/nt, where n = nr of moles of one of the molecules present in the mixture and nt= nr of total moles presents in the misture. With this expression you can calculate the molar fraction of ammonia (the total of it, with means you have "fraction that was absorved + fraction that passed with the stream"). Well after that its simple math and you get your answer.

The best of luck.

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