# Moles and gas volumes HELP!!!!

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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

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

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#2

(Original post by

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

**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

C

_{x}H

_{y}

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(Original post by

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

C

**MexicanKeith**)I would start by writing out the formula for combustion of the hydrocarbon

C

_{x}H_{y}
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**MexicanKeith**)

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

C

_{x}H

_{y}

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#5

(Original post by

CxHy+O2->H2O +CO2

**brookleahy**)CxHy+O2->H2O +CO2

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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(Original post by

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

**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

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#7

(Original post by

So after you've balanced it is that the answer because you have the hydrocarbon formula?

**brookleahy**)So after you've balanced it is that the answer because you have the hydrocarbon formula?

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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(Original post by

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

**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

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#9

(Original post by

How would I know which is the correct balanced equation if you can balance it 2 ways?

**brookleahy**)How would I know which is the correct balanced equation if you can balance it 2 ways?

the equation you need to balance is the general one with x's an y's in it

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(Original post by

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

**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

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#11

(Original post by

Ok thankyou so when you balance CxHy, does that mean you'll have one carbon and one hydrogen?

**brookleahy**)Ok thankyou so when you balance CxHy, does that mean you'll have one carbon and one hydrogen?

so overall

CxHy + (x+y/4) O2 ----> xCO2 + y/2 H2O

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#13

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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(Original post by

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?

**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?

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**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?

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#16

**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

**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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(Original post by

Please be

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

As for finding

**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.
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#18

(Original post by

Ok thankyou but how do you actually find Y and x after completing the equations?

**brookleahy**)Ok thankyou but how do you actually find Y and x after completing the equations?

**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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(Original post by

An example to find

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

y = 0.005 / 0.000417 = 12, therefore

Hopefully you can use a similar method to find

**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**.
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**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**.

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X

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