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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 ml

2)volume of CH4 is 5 ml

3)percentage of CH4 is 50%

4)percentage of C2h4 IS 33.33%

Soumya Sen , 12 Years ago
Grade 12
anser 2 Answers
Aman Bansal

Last Activity: 12 Years ago

Dear Soumya,

Clearly, 

percentage of CH4 is 50%

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Raghav Dhupar

Last Activity: 10 Years ago

Your answer is wrong ,aman 
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 ;-) 
 
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