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Topic: hydrocarbon combustion analysis  (Read 5471 times)

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Offline reema.rifkhan

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hydrocarbon combustion analysis
« on: May 31, 2015, 12:28:40 PM »
ummm....Im new here and I need an answer to this question asap plz... I have no idea on how to do this! :-[

10 cm3 of a gaseous hydrocarbon, CxHy, was mixed with excess oxygen and ignited.
The total gas volume was measured at room temperature and pressure before and
after combustion, and it was found that it had contracted by 20 cm3. On shaking
the remaining gases with excess potassium hydroxide solution, the total gas volume
contracted by a further 40 cm3.
The equation for the complete combustion of CxHy is
CxHy + (x + (y/4))O2  :rarrow: xCO2 + (y/2)H2O

Calculate the molecular formula of CxHy.

The answer is c4 H4 but I can't understand how to get this.
« Last Edit: May 31, 2015, 12:42:54 PM by reema.rifkhan »

Online Hunter2

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Re: hydrocarbon combustion analysis
« Reply #1 on: May 31, 2015, 12:33:56 PM »
What is your atempt. See forum rules.

One hint: the volumes guide to moles.

Offline reema.rifkhan

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Re: hydrocarbon combustion analysis
« Reply #2 on: May 31, 2015, 12:45:38 PM »
What is your atempt. See forum rules.

One hint: the volumes guide to moles.
Well, i don't know how to do this question at all so I didn't include any attempts...
please help if you know how to do this

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Re: hydrocarbon combustion analysis
« Reply #3 on: May 31, 2015, 12:52:30 PM »
Calculate how much mole is 10 cm³ gas. And also you can calculate how much is the 40 cm³ absorbed by KOH. This gives you the moles of CO2 ==> guides the moles of C. The 20 cm³ gives you the values for water.

Now its your turn.

Offline reema.rifkhan

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Re: hydrocarbon combustion analysis
« Reply #4 on: May 31, 2015, 01:07:32 PM »
Calculate how much mole is 10 cm³ gas. And also you can calculate how much is the 40 cm³ absorbed by KOH. This gives you the moles of CO2 ==> guides the moles of C. The 20 cm³ gives you the values for water.

Now its your turn.
10  cm3 gas=>no of CxHymoles= 1/2400 moles
40 cm3 gas => no of CO2 moles=X=1/600 moles
no of H2O moles= y/2=1/1200 moles
so y=1/600 moles
itz like CxHy= C1/600 H1/600 moles
how can I derive the answer (C4 H4) from this?
what can I do next?

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Re: hydrocarbon combustion analysis
« Reply #5 on: May 31, 2015, 02:18:56 PM »
The  10 cm³ gives you the 1/2400 moles. The 40 cm³ gives you 1/600 mole CO2 what means 1/600 moles C so it means you get from 1/2400 mole hydrocarbon 1/600 moles C so it means you have 4 C in the molecule.
The 20 cm³ water steam gives 1/1200 mole. Water contain 2 hydrogen so you have 1/600 mole Hydrogen. So the result for hydrogen is also 4.

The result is  C4H4

Offline reema.rifkhan

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Re: hydrocarbon combustion analysis
« Reply #6 on: May 31, 2015, 02:32:14 PM »
The  10 cm³ gives you the 1/2400 moles. The 40 cm³ gives you 1/600 mole CO2 what means 1/600 moles C so it means you get from 1/2400 mole hydrocarbon 1/600 moles C so it means you have 4 C in the molecule.
The 20 cm³ water steam gives 1/1200 mole. Water contain 2 hydrogen so you have 1/600 mole Hydrogen. So the result for hydrogen is also 4.

The result is  C4H4
I understood it...Thanks a lot... :) :) :)

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