Ideal Gas Equation

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emmald583
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#1
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#1
Hi all,

The last question of a worksheet on the ideal gas equation is confusing me a bit. The question reads:

"20cm3 of a gaseous alkene is exploded with 150cm3 of oxygen. The gases produced in this reaction occupied 110cm3. When these gases were shaken with aqueous sodium hydroxide to remove any excess carbon dioxide remaining after the reaction, the final volume was 30cm3. (All volumes were measured at the same temperature and pressure, and all water produced in the reaction is assumed to be in the liquid phase)."

The first part of the question was to write an equation for the reaction of carbon dioxide with sodium hydroxide, which I believe is CO2 (g) + 2NaOH (aq) -> Na2CO3 (aq) + H2O (l).

However, the second part is where I have run into problems - it asks for the formula of the alkene. Since I know that an alkene burned in oxygen produces carbon dioxide and water, I worked out that the volume of water produced is 60cm3 (as 20+150 is 170, but only 110cm3 of gas was produced, and any water produced was liquid). I have also worked out that, since 110cm3 of gas was produced, and the final volume was 30cm3, 80cm3 CO2 was removed by the sodium hydroxide. Since the only gases (apart from the alkene) involved in the equation are carbon dioxide and oxygen, the 30cm3 remaining must be excess oxygen.

I'm not sure where to go from this information though? Any advice greatfully received!
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MexicanKeith
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(Original post by emmald583)
Hi all,

The last question of a worksheet on the ideal gas equation is confusing me a bit. The question reads:

"20cm3 of a gaseous alkene is exploded with 150cm3 of oxygen. The gases produced in this reaction occupied 110cm3. When these gases were shaken with aqueous sodium hydroxide to remove any excess carbon dioxide remaining after the reaction, the final volume was 30cm3. (All volumes were measured at the same temperature and pressure, and all water produced in the reaction is assumed to be in the liquid phase)."

The first part of the question was to write an equation for the reaction of carbon dioxide with sodium hydroxide, which I believe is CO2 (g) + 2NaOH (aq) -> Na2CO3 (aq) + H2O (l).

However, the second part is where I have run into problems - it asks for the formula of the alkene. Since I know that an alkene burned in oxygen produces carbon dioxide and water, I worked out that the volume of water produced is 60cm3 (as 20+150 is 170, but only 110cm3 of gas was produced, and any water produced was liquid). I have also worked out that, since 110cm3 of gas was produced, and the final volume was 30cm3, 80cm3 CO2 was removed by the sodium hydroxide. Since the only gases (apart from the alkene) involved in the equation are carbon dioxide and oxygen, the 30cm3 remaining must be excess oxygen.

I'm not sure where to go from this information though? Any advice greatfully received!
You're along the right track thinking that the 30cm3 remaining is excess Oxygen, if you write a general formula for an alkene with one double bond then you should be able to write an equation for its combustion. See if you can do it from that information, if not im happy to give you a bit more help!
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emmald583
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(Original post by MexicanKeith)
You're along the right track thinking that the 30cm3 remaining is excess Oxygen, if you write a general formula for an alkene with one double bond then you should be able to write an equation for its combustion. See if you can do it from that information, if not im happy to give you a bit more help!
The general formula for the alkene would be CnH2n, so the unbalanced equation would be CnH2n + O2 -> CO2 + H2O. I'm still having trouble with relating this to the volumes I've worked out, and those given in the question?
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MexicanKeith
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(Original post by emmald583)
The general formula for the alkene would be CnH2n, so the unbalanced equation would be CnH2n + O2 -> CO2 + H2O. I'm still having trouble with relating this to the volumes I've worked out, and those given in the question?
So the first thing to do is balance the equation!

it will look something like this

CnH2n + 1.5n O2 ----> nCO2 + nH2O

make sure you can see why that is balanced!

The next thing to remember is that, at constant pressure and temperature, volume and moles are proportional to one another! pV=nRT so the numbers simply mean, there is initially 6 times as much of the oxygen as there is of the alkene
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emmald583
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So n=4?

Thanks so much for your help, it didn't occur to me to balance the others with ns!
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MexicanKeith
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(Original post by emmald583)
So n=4?

Thanks so much for your help, it didn't occur to me to balance the others with ns!
Yes! n=4

I imagine being the last question on the sheet it's there to challenge you! Well done on understanding it, presuming you're doing A levels that's a pretty tough question!
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Gunjo
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(Original post by MexicanKeith)
So the first thing to do is balance the equation!

it will look something like this

CnH2n + 1.5n O2 ----> nCO2 + nH2O

make sure you can see why that is balanced!

The next thing to remember is that, at constant pressure and temperature, volume and moles are proportional to one another! pV=nRT so the numbers simply mean, there is initially 6 times as much of the oxygen as there is of the alkene
Hi, how did you know it was 6 times as much oxygen?

The way I did it was dividing 80/20 = 4 so x is 4
As CnH2n, it was therefore C4H8.

But I'm interested in your method, please explain?
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MexicanKeith
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(Original post by Gunjo)
Hi, how did you know it was 6 times as much oxygen?

The way I did it was dividing 80/20 = 4 so x is 4
As CnH2n, it was therefore C4H8.

But I'm interested in your method, please explain?
30cm^3 of Oxygen remaining at the end of the reaction, therefore, of the 150cm^3 initially present 120cm^3 must have reacted

120cm^3 of O2 reacting with 20cm^3 of alkene
120/20 is a 6:1 ratio
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Gunjo
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Thanks
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studyfunstudy
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#10
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Hi, I've also been doing these questions and was wondering if anyone has any questions similar so I can get some more practice? thank you
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