# combustion of fuel problem

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octane is burnt with 80% air

if it is theoritical air ,the equation wiill be

C8H18+a(O2+3.76N2)------>xCO2+yH2O+a*(3.76N2)

and form that equation we can find the coefficient a,x,y.

for x=8;

2y=18, so y=9;

2a=2x+y, so 2a=2*8+9, a=12.5

but when it comes to less than theoritical air there should be co present ,shouldn't it?

then what will be the equation

c8h8+a(o2+3.76n2)----->xco2+yh2o+a*(3.76n2)+co

if i give some coefficient to co my question is how i can find the value coefficient of co here ?

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why don't you do it like the first one?

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i tried .lets see...

if we declare another coefficient d for co then

c8h8+a(o2+3.76n2)----->xco2+yh2o+a*(3.76n2)+dco

8=x+d;..........(1)

2y=18, so y=9;

2a=2x+y+d

=>2x+d=2a-9;...........(2)

we get 3 variable and two equation.but we need 3 equations to solve for 3 variable...

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if you're sure with the equations then let's try...

First having equations less than the number of variables doesn't mean there is no solution.

I guess you should write d or x interms of 'a' then insert any number that satisfy the equation.

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then how can i do it

i cant proceed any further than the equations i wrote above

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