Alkanes

36. ​​ O/N​​ 2013/P12/Q11

The complete combustion of 20cm3​​ ​​ of a gaseous alkane, X, requires 130​​ cm3​​ ​​ of oxygen. Both​​ volumes were measured at r.t.p..

What could be the identity of X?

A​​ Butane

B​​ Ethane

C​​ Methane

D​​ Propane

 

Answer: ​​ A

Explaination

First,​​ we need to recall moles, as we know that gas having same moles will have same volume at r.t.p with irrespective of any gas.

For e.g:  ​​ ​​​​ 1 mole of nitrogen gas = 24dm3 volume

If there is 2 moles of nitrogen gas = 48dm3 volume (volume increase with respect to mole)​​ 

If there is 3 moles of nitrogen gas = 72 volume (3 × 24)

So let’s have a look for question:

Alkane (X) + O2​​ ---------------> CO2​​ + H2O  ​​ ​​​​ (alkane combustion)

 ​​ ​​​​ 

To find the simplest whole ratio

Alkane X : Oxygen 

20cm320 :130cm320

1 :6.5

Now to get this above ratio we need to find out gas X (butane/propane/ethane/methane) which combust at this ratio i.e: 1:6.5

Hence,​​ butane​​ is the gas​​ which combust at 1 : 6.5 ratio.