# Moles and gas volumes HELP!!!!

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#1
Question:
10cm3 of a hydrocarbon of formula CxHy was burnt in excess oxygen. At the end of the reaction 0.045g of liquid water had been formed and the total volume of gas left was 80cm3, of which 30cm3 was unburnt oxygen. Deduce the values of x and y in the formula for the hydrocarbon.

I know how to do the Moles=mass/Mr equation and n=v/24 but I still have no idea how you get the values of x and y- I've tried so many different ways! Please can someone give me some guidance?

Thankyou
0
3 years ago
#2
(Original post by brookleahy)
Question:
10cm3 of a hydrocarbon of formula CxHy was burnt in excess oxygen. At the end of the reaction 0.045g of liquid water had been formed and the total volume of gas left was 80cm3, of which 30cm3 was unburnt oxygen. Deduce the values of x and y in the formula for the hydrocarbon.

I know how to do the Moles=mass/Mr equation and n=v/24 but I still have no idea how you get the values of x and y- I've tried so many different ways! Please can someone give me some guidance?

Thankyou
I would start by writing out the formula for combustion of the hydrocarbon

CxHy
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#3
(Original post by MexicanKeith)
I would start by writing out the formula for combustion of the hydrocarbon

CxHy
CxHy+O2->H2O +CO2
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#4
(Original post by MexicanKeith)
I would start by writing out the formula for combustion of the hydrocarbon

CxHy
But after you've done the formula and worked out moles and volume how do you find x an y- do you have to reearrange an equation?
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3 years ago
#5
(Original post by brookleahy)
CxHy+O2->H2O +CO2
Equation has to be balanced

eg

CH4 + 2O2 --> CO2 + 2H2O

or

C5H10 + 7.5 O2 ---> 5CO2 + 5H2O

CxHy is a more general case so the numbers will all be in terms of x and y
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#6
(Original post by MexicanKeith)
Equation has to be balanced

eg

CH4 + 2O2 --> CO2 + 2H2O

or

C5H10 + 7.5 O2 ---> 5CO2 + 5H2O

CxHy is a more general case to the numbers will all be in terms of x and y
So after you've balanced it is that the answer because you have the hydrocarbon formula?
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3 years ago
#7
(Original post by brookleahy)
So after you've balanced it is that the answer because you have the hydrocarbon formula?
no, when you balance the equation

CxHy + [?] O2 ----> [?] CO2 + [?] H2O

you are then in a position to use mole calculations and ideal gas equation to work out the answer.

so first, just try to balance it
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#8
(Original post by MexicanKeith)
Equation has to be balanced

eg

CH4 + 2O2 --> CO2 + 2H2O

or

C5H10 + 7.5 O2 ---> 5CO2 + 5H2O

CxHy is a more general case so the numbers will all be in terms of x and y
How would I know which is the correct balanced equation if you can balance it 2 ways?
0
3 years ago
#9
(Original post by brookleahy)
How would I know which is the correct balanced equation if you can balance it 2 ways?
The two equations i gave were just examples of balanced combustion equations

the equation you need to balance is the general one with x's an y's in it 0
#10
(Original post by MexicanKeith)
The two equations i gave were just examples of balanced combustion equations

the equation you need to balance is the general one with x's an y's in it Ok thankyou so when you balance CxHy, does that mean you'll have one carbon and one hydrogen?
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3 years ago
#11
(Original post by brookleahy)
Ok thankyou so when you balance CxHy, does that mean you'll have one carbon and one hydrogen?
so if I balance that in general I have x carbons and y hydrogens

so overall

CxHy + (x+y/4) O2 ----> xCO2 + y/2 H2O
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#12
Would you say this is correct:

C4H8 + 6O2 -> 4CO2+ 4H2O
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3 years ago
#13
(Original post by brookleahy)
Would you say this is correct:

C4H8 + 6O2 -> 4CO2+ 4H2O
Im afraid not

you should know that any gas occupies the same volume per mole

use the information you're given to work out the volume of CO2 produced in the reaction, what do you get?
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#14
(Original post by MexicanKeith)
Im afraid not

you should know that any gas occupies the same volume per mole

use the information you're given to work out the volume of CO2 produced in the reaction, what do you get?
So the volume is 0.05dm3?
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#15
(Original post by MexicanKeith)
Im afraid not

you should know that any gas occupies the same volume per mole

use the information you're given to work out the volume of CO2 produced in the reaction, what do you get?
So my equation still isn't balanced?
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3 years ago
#16
(Original post by brookleahy)
Question:
10cm3 of a hydrocarbon of formula CxHy was burnt in excess oxygen. At the end of the reaction 0.045g of liquid water had been formed and the total volume of gas left was 80cm3, of which 30cm3 was unburnt oxygen. Deduce the values of x and y in the formula for the hydrocarbon.

I know how to do the Moles=mass/Mr equation and n=v/24 but I still have no idea how you get the values of x and y- I've tried so many different ways! Please can someone give me some guidance?

Thankyou
Please be incredibly careful when using the n=v/24 formula. This only applies if v is measured in dm^3. You're using cm^3 which are 1,000 times smaller. Convert to dm^3 by dividing by 1,000.

First, use the n=v/24 formula to find out how many moles of hydrocarbon there are (assume it's a gas). Then find out how many moles of liquid water you have using the moles = mass/Mr equation. Since there are no hydrogen atoms involved anywhere else in the reaction, you should be able to find y from this.

As for finding x, think about what the other 50cm^3 of gas must be. Complete combustion of a hydrocarbon creates water and what else? Find out how many moles of this gas there are. Since this is where all the carbon has gone, you can find x from this.
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#17
(Original post by TheMindGarage)
Please be incredibly careful when using the n=v/24 formula. This only applies if v is measured in dm^3. You're using cm^3 which are 1,000 times smaller. Convert to dm^3 by dividing by 1,000.

First, use the n=v/24 formula to find out how many moles of hydrocarbon there are (assume it's a gas). Then find out how many moles of liquid water you have using the moles = mass/Mr equation. Since there are no hydrogen atoms involved anywhere else in the reaction, you should be able to find y from this.

As for finding x, think about what the other 50cm^3 of gas must be. Complete combustion of a hydrocarbon creates water and what else? Find out how many moles of this gas there are. Since this is where all the carbon has gone, you can find x from this.
Ok thankyou but how do you actually find Y and x after completing the equations?
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3 years ago
#18
(Original post by brookleahy)
Ok thankyou but how do you actually find Y and x after completing the equations?
An example to find y:

10cm^3 of hydrocarbon = 0.01dm^3. n=0.01/24 = 0.000417 moles of hydrocarbon
0.045g / 18 = 0.0025 moles of water
0.0025 moles of water contains 0.005 moles of hydrogen atoms because there are 2 hydrogen atoms per water molecule.
y = 0.005 / 0.000417 = 12, therefore y = 12

Hopefully you can use a similar method to find x.
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#19
(Original post by TheMindGarage)
An example to find y:

10cm^3 of hydrocarbon = 0.01dm^3. n=0.01/24 = 0.000417 moles of hydrocarbon
0.045g / 18 = 0.0025 moles of water
0.0025 moles of water contains 0.005 moles of hydrogen atoms because there are 2 hydrogen atoms per water molecule.
y = 0.005 / 0.000417 = 12, therefore y = 12

Hopefully you can use a similar method to find x.
Thankyou so much!!
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#20
(Original post by TheMindGarage)
An example to find y:

10cm^3 of hydrocarbon = 0.01dm^3. n=0.01/24 = 0.000417 moles of hydrocarbon
0.045g / 18 = 0.0025 moles of water
0.0025 moles of water contains 0.005 moles of hydrogen atoms because there are 2 hydrogen atoms per water molecule.
y = 0.005 / 0.000417 = 12, therefore y = 12

Hopefully you can use a similar method to find x.
So would x=5 to make C5H12 Pentane? I worked this out with my own knowledge rather than the method please could advise me which numbers to use to get 5?
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