How do I balance this combustion reaction?

Question

I have butene being burned in a fuel rich engine and I can't seem to balance this reaction

$C_{4}H_{8}+a(O_{2}+3.76N_{2})->bCO_{2}+cH_{2}O+dC_{4}H_{8}+e(3.76N_{2})$

When I try write equations for the elements, I get 5 equations and 4 unknowns which can't be solved. Here are my equations:

Carbon: $4=b+4d$

Oxygen: $2a=2b+c$

Nitrogen: $3.76a=3.76e;\quad a=e$

Hydrogen: $2a=2c+8d$

Edit: I'm not sure if this makes a different but it says that water in the combustion products is removed before the dry analysis of the exhaust. The DRY analysis has a composition of 14.95% $CO_{2}$, 0.75% $C_{4}H_{8}$, 0% $CO$, 0% $H_{2}$, 0% $O_{2}$, and balance $N_{2}$ (I don't know what that means)

Answer 1

Balanced equation for the combustion of $\ce{C4H8}$ is $$\ce{C4H8 + 6O2 -> 4CO2 + 4H2O}$$ Nitrogen is included only to indicate that it is present in the medium with some specific ratio to the oxygen amount --well technically it is included to make question look complicated.

$\ce{C4H8}$ in the products side is included to indicate that not all of it can be combusted but some escapes. And it really is arbitrary, depends on the the engineering of combustion room etc -which is why you have 4 equations, 5 unknowns: d is free to vary. But you are given the dry analysis to fix its value.

Looking at the equations given, you have

b = c = 4(1-d)
a = e = b + c/2 = 6(1-d) = 3b/2

And dry analysis says ratio of $\ce{CO2}$:$\ce{C4H8}$ is 14.95:0.75 = 19.93 (b/d) after the combustion. And this is the last equation you need.

Answer 2

See this video for a complete description of a matrix-based approach. It starts by showing how to set up the matrix. It then takes each step of reducing the matrix and extracting coefficients.