got it ! IIT JEE 1993(the toughest one )

here we go..

CH₄+2O₂ = CO2+2H2O

from here we see that for each volume(V) of CH4 we should supply 2V O2 "theoriticaly" but 3*2V=6V practically for the complete comb.

 

now

we know that Heat generated per sec.   (H comb)X (gas supplied per sec )

=k (H comb)X (gas supplied per sec )

for the two gas CH4 n C4H10 "Heat generated per sec" is same hence we get

(H comb of CH4)X (CH45 supplied per sec )=(H comb of C4H10)X (C4H10 supplied per sec )

from above we get

Butane supplied per sec = x*809/2978---------------------------(1)

 

now for combstion of butane we have

C4H10+13/2O2 = 4CO2+5H2O

ie. we should supply 13/2V oxygen for each V of Butane

and thus practicaly 3*13/2V of oxygen

but from (1) we have V= x*809/2978

thus rate of supply of oxygen =3*x*809/2978

 

calculating we get

Butane supplied per sec =0.271x L /sec

and Oxygen=13/2*0.271 =5.29x L/sec

 

I must say I've done this before and hope this is right again 

m i rit ?

(typed too much ? hena ?)