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Topic: Mixture combustion (tough)  (Read 4189 times)

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Offline Rutherford

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Mixture combustion (tough)
« on: January 14, 2013, 11:42:36 AM »
A mixture of substance with the formula C3H6O and C3H8O is burned in the presence of a stoichiometric amount of air (20% oxygen and 80% of nitrogen by volume) in an eudiometer to produce a mixture of gasses with a total volume of 5.432dm3 at STP. The mixture is then bubbled through Ba(OH)2 solution, its volume decreases by 15.46%. Calculate the molar ratio of the substances in the beginning mixture.

Attempt:
xC3H6O+4.5xO2 :rarrow: 3xCO2+3xH2O
yC3H8O+5yO2 :rarrow: 3yCO2+4yH2O
I mark the number of moles of C3H6O with x and of C3H8O with y.

The volume decrease is because of CO2 whose number of moles is n=5.432*0.1546/22.4=0.0375mole, this means that: x+y=0.0375

The remaining mixture should contain nitrogen (V=4.592dm3) so the volume of oxygen that was spent was 4 times smaller and its number of moles is: n=4.592/(4*22.4)=0.0512mole, from the reaction equation: 4.5x+5y=0.0512. When this is solved with the first equation (x+y=0.0375) I get a negative answer.

Then I thought that maybe water was in gas state, too. The first equation is the same, now to the second one. The number of moles of the mixture is equal to the sum of the moles of water and 4*number of moles of oxygen (because there was 4 times more nitrogen in air than oxygen): 4.592/22.4=3x+4y+4*(4.5x+5y), i.e. 21x+24y=0.205. When solved with equation x+y=0.0375 again a negative answer is obtained, so I need help with this.

Offline Borek

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Re: Mixture combustion (tough)
« Reply #1 on: January 14, 2013, 12:22:07 PM »
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Offline Rutherford

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Re: Mixture combustion (tough)
« Reply #2 on: January 14, 2013, 12:42:00 PM »
Oh there is the mistake. Thanks. It should be 3x+3y=0.0375, therefore x+y=0.0125, but again I get a negative answer in both ways  :-\.

Offline DrCMS

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Re: Mixture combustion (tough)
« Reply #3 on: January 14, 2013, 04:48:47 PM »
Where did the water produced by the combustion go?

Offline Rutherford

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Re: Mixture combustion (tough)
« Reply #4 on: January 15, 2013, 05:33:19 AM »
I don't understand your question. First I thought that it isn't in gas state, but in the problem it wasn't said that the mixture is returned to the beginning temperature (before the combustion) so I calculated then when water is in gas state. I get negative answers.

Offline Dan

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Re: Mixture combustion (tough)
« Reply #5 on: January 15, 2013, 07:33:26 AM »
xC3H6O+4.5xO2 :rarrow: 3xCO2+3xH2O
yC3H8O+5yO2 :rarrow: 3yCO2+4yH2O

These equations are not balanced.

First one: LHS has 10 O; RHS has 9 O
Second: LHS has 11 O; RHS has 10 O

Fix this and you should get +ve solutions.
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Offline Rutherford

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Re: Mixture combustion (tough)
« Reply #6 on: January 15, 2013, 08:11:43 AM »
I forgot the oxygen in the molecules that are combusted  :-[. It's:
xC3H6O+4xO2 :rarrow: 3xCO2+3xH2O
xC3H8O+4.5xO2 :rarrow: 3xCO2+4xH2O
I got the answer when assuming that water vapors condensed, but it isn't said in the problem that it condensed. That's a little issue when solving this type of problems, sometimes it is included, sometimes not. Anyway the answer is 4:1 and I got it 4.2:1. I think its okay. Thanks very much for the correction.

Offline AWK

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Re: Mixture combustion (tough)
« Reply #7 on: January 15, 2013, 08:21:59 AM »
At STP water is practically removed. What is a vapor pressure of water at 273 K?
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Offline curiouscat

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Re: Mixture combustion (tough)
« Reply #8 on: January 15, 2013, 08:26:54 AM »
At STP water is practically removed. What is a vapor pressure of water at 273 K?

Is it saying the reaction was at STP or merely reporting volumes at STP?

Offline AWK

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Re: Mixture combustion (tough)
« Reply #9 on: January 15, 2013, 08:48:52 AM »
Hence my the second sentence (question).
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Offline Borek

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Re: Mixture combustion (tough)
« Reply #10 on: January 15, 2013, 09:03:57 AM »
mixture of gasses with a total volume of 5.432dm3 at STP

I would say volumes measured at STP - so water is mostly in a liquid form.
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Offline Rutherford

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Re: Mixture combustion (tough)
« Reply #11 on: January 16, 2013, 06:28:46 AM »
mixture of gasses with a total volume of 5.432dm3 at STP

I would say volumes measured at STP - so water is mostly in a liquid form.
When just looking at the sentence I think it could be both, but maybe the sentence you wrote seems more probably. It is a little ambiguous formulated.

Offline AWK

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Re: Mixture combustion (tough)
« Reply #12 on: January 16, 2013, 07:47:15 AM »
In all problems at STP water is removed as a default. For very precise calculations you should take into account a vapor pressure at 273 K which is 4.6 mmHg, hence neglecting water pressure gives 0.6 % error.
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