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Physics, Chemistry and Biology: Post your doubts here!

Hello090078601

Hello090078601

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How to solve:

20 cm^3 of a hydrocarbon was burnt in 175 cm^3 of oxygen. After cooling, the volume of the remaining gases was 125 cm^3. The addition of aqueous sodium hydroxide removed carbon dioxide leaving 25 cm^3 of unreacted oxygen.


iii) Deduce the formula of the hydrocarbon and the balanced equation for the reaction.

Why does it say in the mark scheme that any combustion equation of an ALKENE would do? Why not an alkane?

http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Chemistry (0620)/0620_s13_qp_33.pdf
 
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Awesome12

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Phoenix Blood said:
View attachment 43337
Awesome12 Xylferion
Click to expand...
Find the moles using the formula
Moles = mass/Mr

Moles of Ca = 0.12
Moles of H2O = 0.2

Now the ratio of Ca: H2O is
1 : 2
so the no. of moles of H2O should be

0.12 * 2 = 0.24

However water has a fewer no. of moles (calculated in first part), so Calcium is in excess

Water has 0.2 moles (only 0.2 moles will react), and so will Calcium (calcium is in excess)

No. of moles remaining (o.24 - 0.2) = 0.04

Mass of calcium =

0.04 * 40
=1.6 g


This is quite complex, do ask again if you don't understand something
 
Phoenix Blood

Phoenix Blood

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Hello090078601 said:
How to solve:

20 cm^3 of a hydrocarbon was burnt in 175 cm^3 of oxygen. After cooling, the volume of the remaining gases was 125 cm^3. The addition of aqueous sodium hydroxide removed carbon dioxide leaving 25 cm^3 of unreacted oxygen.

i) Volume of oxygen used: ___________

ii) Volume of carbondioxide formed: __________

iii) Deduce the formula of the hydrocarbon and the balanced equation for the reaction.
Click to expand...
Awesome12, lol, plll help
 
Xylferion

Xylferion

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Phoenix Blood said:
View attachment 43335

'ElllpP!

Xylferion Awesome12
Click to expand...

C1V1 = C2V2 , remember this formula.

Concentration 1 is given to you as 2.20 dm^3.
Volume 1 is 20 cm^3.

convert the cm^3 to dm^3. /1000

So volume 1 = 0.02 dm^2.

They want you to find the concentration of lithium hydroxide.

So let concentration 2 be "x".

Volume 2 = 25 cm^3. Convert it to dm^3. You get 0.025 dm^3.

So... 2.20 * 0.02 = 0.025 * x

x = 1.76 mol / dm^3.
 
Phoenix Blood

Phoenix Blood

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Awesome12 said:
Find the moles using the formula
Moles = mass/Mr

Moles of Ca = 0.12
Moles of H2O = 0.2

Now the ratio of Ca: H2O is
1 : 2
so the no. of moles of H2O should be

0.12 * 2 = 0.24

However water has a fewer no. of moles (calculated in first part), so Calcium is in excess

Water has 0.2 moles (only 0.2 moles will react), and so will Calcium (calcium is in excess)

No. of moles remaining (o.24 - 0.2) = 0.04

Mass of calcium =

0.04 * 40
=1.6 g


This is quite complex, do ask again if you don't understand something
Click to expand...
Mass of Ca is 0.08 :(
 
Xylferion

Xylferion

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Phoenix Blood said:
Awesome12, lol, plll help
Click to expand...

First write the equation.

CxHy + O2 = CO2 + H2O

Volume of oxygen used = will be total oxygen used - oxygen not used.

175 cm^3 used, 25 not used so 175-25 = 150 cm^3.

Volume of carbon dioxide formed = volume of remaining gases - volume of oxygen unreacted
= 100 cm^3.

So your data now is.

20 cm^3 of the hydrocarbon reacted with 150 cm^3 of oxygen to form 100 cm^3 of Carbon dioxide.

The ratio is 2:15:10

Using this ratio you can find the number of moles of H2O that forms.

If the ratio of CO2 to H2O is 1:1.

15 moles produce 10 moles of CO2. Then 15 moles will also produce 10 moles of H20.

Your formula should look like this now.

2CxHy + 15O2 ------> 10CO2 + 10H2O.

there are 10 carbons on the left hand side.

So there must be 10 carbons on the right hand side, but there are 2 moles of the hydrocarbon so it will be 5 carbons on the left.

2C5Hy + 15O2 ------> 10CO2 + 10H2O.

There are 20 hydrogens on the right.

So there has to be 20 on the left.

but 2 moles of the hydrocarbon, makes it 10.

So final formula = C5H10 and final equation = 2C5H10 + 1502 -----> 10CO2 + 10H20.
 
Phoenix Blood

Phoenix Blood

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Xylferion said:
First write the equation.

CxHy + O2 = CO2 + H2O

Volume of oxygen used = will be total oxygen used - oxygen not used.

175 cm^3 used, 25 not used so 175-25 = 150 cm^3.

Volume of carbon dioxide formed = volume of remaining gases - volume of oxygen unreacted
= 100 cm^3.

So your data now is.

20 cm^3 of the hydrocarbon reacted with 150 cm^3 of oxygen to form 100 cm^3 of Carbon dioxide.

The ratio is 2:15:10

Using this ratio you can find the number of moles of H2O that forms.

If the ratio of CO2 to H2O is 1:1.

15 moles produce 10 moles of CO2. Then 15 moles will also produce 10 moles of H20.

Your formula should look like this now.

2CxHy + 15O2 ------> 10CO2 + 10H2O.

there are 10 carbons on the left hand side.

So there must be 10 carbons on the right hand side, but there are 2 moles of the hydrocarbon so it will be 5 carbons on the left.

2C5Hy + 15O2 ------> 10CO2 + 10H2O.

There are 20 hydrogens on the right.

So there has to be 20 on the left.

but 2 moles of the hydrocarbon, makes it 10.

So final formula = C5H10 and final equation = 2C5H10 + 1502 -----> 10CO2 + 10H20.
Click to expand...
Hello090078601
 
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ChubbyBunny

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Guys I have a lot of problem in organic chemistry specially the ester , carboxylic acid part and further ..got any notes for it ?
 
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