Ten millilitre of a mixture of `CO,CH_4`, and `N_2` exploded with an excess of `O_2` and gave a contaction of `6.5ml`. When the residual gas was treated with `NaOH`, there was further contraction of 7 ml. What is the composition of the original mixture?

Explanation

Let he volum of `CO` be `xml`
`CH_4=yml`
`N_2=(10-x-y)ml`
The explosion reaction is `(N_2` does not take part in the reaction)
`CO+(1)/(2)O_2rarrunderset(xml)CO_2` ..(1)
`underset(yml)CH_4+underset(2yml)2O_2rarrunderset(yml)CO_2+underset(-)2H_2O` ..(2)
volume of `CO_2=x+y=7ml` .(3)
Since the water vapour condenses to practically zero volume of water, the decrease in the volume of cooling (i.e., contraction of the volume) is the volume of water vapour.
Volume of reactant`=`volume of product `+` contraction
'`Vco +VcH_4+V_(O_2)(excess)+cancelV_(N_2)=V_(CO_2)+V_(O_2)(l eft)+cancelV_(N_2)`
`cancelx+cancely+V_(O_2)(excess)=cancelx+cccancely+O_2(l eft)+6.5`
`V_(O_2)(excess)-V_(O_2)(l eft)=V_(O_2)(used)=6.5ml`
`V_(O_2)(used)=6.5ml`
from equation (1) and (2),
`V_(O_2)(used)=((x)/(2)+2y)ml=6.5ml`
`V_(O_2)(used)=4y+x-13`.. (4)
From equation (3) and (4) solve for x and y.
`x=` volume of `CO=5ml`
`y=`volume of `CH_4=2ml`
volume of `N_2=(10-5-2)3ml`