# 9ml of a mixture of methane and ethylene was exoloded  with 30ml oxygen.after cooling the residual gas mixture was found to be 21 ml on treatment with NAOH solution,it was reduced to 7 ml.which is true given mixture?1)volume of CH4 is 4 ml2)volume of CH4 is 5 ml3)percentage of CH4 is 50%4)percentage of C2h4 IS 33.33%

Aman Bansal
592 Points
11 years ago

Dear Soumya,

Clearly,

 percentage of CH4 is 50%

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Thanks

Aman Bansal

Raghav Dhupar
24 Points
8 years ago
The answer is 1) vol. Of CH4 is 4 ml
Sol.- let vol of CH4 be "a" and C2H4 be "b" .
As oxygen is in exess
Therefore reaction of methane and ethane write as
CH4 + 2O2 ---> CO2 + 2H2O liquid(as cooled)
a           2a            a          ----        (these are moles)

And of ethylene as
C2H4 + 3O2 ------> 2CO2   +   2H2O
b            3b                   2b              ----

As after cooloing mixture was found 21 ml (thats the water neglected)
And it was teduced to 7 on treatment with NaOH therefore CO2 consumed by NaOH is 14 ml
And equating moles of CO2 as
a+ 2b = 14-------(1)
And ,  we know that a+b=9 ------(2)
And (1)-(2)
b = 5 ,
Therefore a= 4 ans.  Hope you satisfy by my answer .
My name is raghav dhupar and im of class +1 ,jalandhar ,punjab .
Thanks and enjoy your studies. And its very diff. To type ll this on smartphone ;-)