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A 20 mL mixture of ethane, ethylene, and...

A 20 mL mixture of ethane, ethylene, and `CO_2` is heated with `O_2`. After the explosion, there was a contraction of 28 mL and after treatment with KOH, there was a further contraction of 30 mL. What is the composition of the mixture?

Text Solution

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Let the volume of ethane, ethylene, and `CO_(2)`, respectively, are a,b and `(20-a-b)mL`
Now contraction after cooling`=28mL`
`thereforeV_(R)-V_(P)=25`
`V_(R)=` Volume of `C_(2)H_(6)+` volume of `C_(2)H_(4)+` volume of `CO_(2)+` volume of `O_(2)` used for combustion
`V_(P)=` volume of `CO_(2)` produced (Volume of `H_(2)O` is not taken since it condenses)
Considering the combustion gases.
(a). `C_(2)H_(6)+(7)/(2)O_(2)to2CO_(2)+3H_(2)O`
(b). `C_(2)H_(4)+3O_(2)to2CO_(2)+2H_(2)O`
(c). `CO_(2)to` no reaction
`V_(R)=(a+(7)/(2)a)+(b+3b)+(20-a-b)`
`V_(P)=(2a+2b)+(20-a-b)`
`V_(P)-V_(R)=(5)/(2)a+2b=28`
`therefore5a+4b=56` ..(i)
There is a further contraction of 32 mL on treatment with KOH. Volume of `CO_(2)` produced `+` Volume of `CO_(2)` original
`=32`
`(2a+2b)+(20-a-b)=32`
`therefore a+b=12` ...(ii)
On solving (i) and (ii) we get
`a=8mL` (volume of `C_(2)H_(6))`
`b=4mL` (Volume of `C_(2)H_(4))`
Volume of `CO_(2)=8mL`
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