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Chemistry: Post your doubts here!

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when 20cm (cube) of a gaseous hydrocarbon were completely burnt in excess of oxygen, 60cm(cube) of carbon dioxide and 40cm(cube) of water vapour were formed, all volumes being measured at same temperature and pressure.

what is the formula of the hydrocarbon ?

A.C2H6
B.C3H4
C.C3H6
D.C3H8

ANS : B

how to work out this question ?

Thanks
 
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when 20cm (cube) of a gaseous hydrocarbon were completely burnt in excess of oxygen, 60cm(cube) of carbon dioxide and 40cm(cube) of water vapour were formed, all volumes being measured at same temperature and pressure.

what is the formula of the hydrocarbon ?

A.C2H6
B.C3H4
C.C3H6
D.C3H8

ANS : B

how to work out this question ?

Thanks
Vol of oxygen : 80cm^3
Hydrocarbon + O2 --------> CO2 + H2O
Set ratio.
1Hydrocaron + 4O2 ---> 3CO2 + 2H2O
You can see there are 3 carbon and 4 hydrogen hence C3H4
 
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help please
Initial moles H2O = 1
Initial moles CO = 1
Eq moles H2O = 1-x
Eq moles CO = 1-x
Eq moles H2 = x
Eq moles CO2 = x

Total moles = (1-x) + (1-x) + x + x = 2
if hydrogen occupies 33% = 2/3 moles hence CO2 is also 2/3 moles and CO and H2O have 1/3 moles

Now put the values into the equilibrium law.
 
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http://www.sheir.org/a-level-chemistry-11-june2011.pdf
Question 8?
And in 11 why's D wrong?
13 My answer's D. What did I do wrong in this calculation:
Al2O3: 6HCL
0.02x6= 2xvolume/1000
AND I DONT GET 15 AND 16 AT ALL! :(
Q13- 4Al + 3O2 ---> 2AlO3
Al2O3 + 6HCl ----> 2AlCl3 + 3H2O

Ratio of Al : Al2O3
4 = 2
0.02 = ?
=0.01 moles of Al2O3
now ratio of Al2O3 : HCl
1 = 6
0.01 = ?
=0.06
use the formula no.of moles = vol x conc. to find the volume
that will be 0.06/2 = 0.03 dm3
 
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Q13- 4Al + 3O2 ---> 2AlO3
Al2O3 + 6HCl ----> 2AlCl3 + 3H2O

Ratio of Al : Al2O3
4 = 2
0.02 = ?
=0.01 moles of Al2O3
now ratio of Al2O3 : HCl
1 = 6
0.01 = ?
=0.06
use the formula no.of moles = vol x conc. to find the volume
that will be 0.06/2 = 0.03 dm3
But why don't you take the moles of Al2O3 to that of HCl directly?
 
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