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Hard Chem MC question

Hydrocarbon Combustion



10 cm3 of a gaseous hydrocarbon was mixed with excess oxygen and ignited. The gas volumes were measured at room temperature and pressure before and after combustion and it was found that the total gas volume had contracted by 20 cm3.
Given that combustion was complete, the formula of the hydrocarbon was
(edited 6 years ago)
Screen Shot 2018-03-22 at 12.33.03 AM.png89.6KB
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Avatar for KaylaB
KaylaB
20
Original post by kashifq
Hydrocarbon Combustion


What exactly is your question?
I make it A

10 C4Hx + n O2 -> 40 CO2 + 5x H2O
oxygens balance
=> 2n = 5x + 80
gas volumes (water is l at std conditions)
10 + n - 20 = 40
=> n = 50
x = 4
Original post by kashifq
Hydrocarbon Combustion



10 cm3 of a gaseous hydrocarbon was mixed with excess oxygen and ignited. The gas volumes were measured at room temperature and pressure before and after combustion and it was found that the total gas volume had contracted by 20 cm3.
Given that combustion was complete, the formula of the hydrocarbon was


Are you sure there wasn't any other options?

Because according to my working out, neither options are correct. As you go down the options from A-> D the answer gets closer to 20 in my working out.
Here is how I did it.

General equation for combustion of the hydrocarbon = C4HX + (4+ 1/4X)O2 -> 4CO2 + X/2 H2O

Now... given that pressure and volume are constant for both readings and use of ideal gas equation (pV = nRT) we can deduce that P IS DIRECTLY PROPORTIONAL TO V.

C4HX + (4+1/4X)O2 -> 4CO2 + X/2H2O difference (product-reactatnst)
mole ratio (x=4) 1 5 4 2 D (difference in volume product-reactant)
volume 10 50 40 20 0

moles R (x=6) 1 5.5 4 3 D
volume 10 55 40 30 +5

moles R ( x=8) 1 6 4 4 D
volume 10 60 40 40 +10

moles R (x=10) 1 6.5 4 5 D
volume cm3 10 65 40 50 +15


now as you can clearly see the difference in volumes increases by 5 cm^3 as we go down each option from option A to D. The difference in volume is what we are trying to get as +20 as thats whats stated in the question.
(edited 6 years ago)
Original post by kashifq
Hydrocarbon Combustion



10 cm3 of a gaseous hydrocarbon was mixed with excess oxygen and ignited. The gas volumes were measured at room temperature and pressure before and after combustion and it was found that the total gas volume had contracted by 20 cm3.
Given that combustion was complete, the formula of the hydrocarbon was


heres a better table of results from my original post.
Untitled.png10.8KB
Reply 5
Avatar for kashifq
kashifq
OP
7
The markscheme says A 🙁, H2O is a liquid at rtp
(edited 6 years ago)
Original post by dip0
heres a better table of results from my original post.


I think it is assuming the water will have cooled into a liquid so A would be the correct answer
Original post by Viren123
I think it is assuming the water will have cooled into a liquid so A would be the correct answer


Yes that makes much more sense now treating volume of liquid water as close to 0cm^3 ( as there are fewer moles of it produced) then now there will be a difference of +20cm^3

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