QUESTION Butene, C4H8 is burned in an engine with a fuel-rich air-fuel ratio. Dry analysis of the exhaust gives the following volume percents: $\ce{CO2} = 14.95%$, $\ce{O2} = 0%$, $\ce{CO} = 0%$, $\ce{H2} = 0%$, $\ce{C4H8} = 0.75%$, with the rest being $\ce{N2}$. Higher heating value of this fuel is 46.9 MJ/kg. Write the balanced chemical equation for one mole of this fuel at these conditions. Calculate the air-fuel ratio, equivalence ratio, lower heating value of the fuel, and energy released when one kg of this fuel is burned in the engine with a combustion efficiency of 98%.
My attempt at solution: I write the stoichiometric equation like so: $$ \ce{C4H8 + \frac{1}{\phi}(4+(8/4))(O2 +3.773N_2)\rightarrow n_1 C4H8 +n_2 CO2 +n_3 H2O}$$ where, $\phi$ is the equivalence ratio(a engineering term used to describe the ratio of $(F/A)_{\text{actual}}$ to $(F/A)_{\text{stoichiometric}}$.
So,now I have 3 unknowns, $n_1$, $n_2$ and $\phi$.
I unfortunately get $n_2=0$ after doing the mole balance of carbons,hydrogen and nitrogen.
I understand that the percentages given by volume are the mass fractions which should later form a separate equation. But, I still get $n_2=0$. Where have I gone wrong?(Should I still consider putting in $\ce{O2}$ and $\ce{CO}$ into the above equation even though their %by volume are zero?)
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to make it look like chemistry. $\endgroup$