10 " mL of " a mixture of methane, ethylene. And carbon dioxide was exploded with excess of air. After the explosion, there was a contraction of 17 mL and after treatment with KOH, there was a further contraction of 14 mL. What was the composition of the 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`