Take the 2-minute tour ×
Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers and students. It's 100% free, no registration required.

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?)

share|improve this question
I have edited your question to make your chemical formulas equations look prettier. When using MathJax, enclosed chemistry in \ce{...} to make it look like chemistry. –  Ben Norris Jul 7 '13 at 18:42
add comment

1 Answer

I see three flaws in your reasoning.

First, butane is $\ce{C4H10}$. The formula $\ce{C4H8}$ refers to several isomers of butene. I will keep using your formula, but you should specify whether you mean butane or $\ce{C4H8}$.

Second, you know the relationship between your variables $n_1$ and $n_2$ because you have the volume percents of $\ce{C4H8}$ and $\ce{CO2}$ in the exhaust. Thus, you have fewer variables.

$$\dfrac{n_1}{n_2} = \dfrac{0.75}{14.95}$$

Third, I am not sure why $\ce{N2}$ is in the equation. It does not participate in the reaction, so it should not be there. Likewise, you do not need to add $\ce{CO}$ and $\ce{H2}$ because you know that none is formed. You can balance the equation and determine the stoichiometric coefficients of $\ce{CO2}$ and $\ce{H2O}$ from the conservation of mass:

$$\ce{C4H8 + 6O2 -> 4CO2 + 4H2O}$$

To include one of your variables, we assume that the reaction does not run to completion, and that some amount $n_1$ of $\ce{C4H8}$ remains unreacted. Since each 1 equivalent of $\ce{C4H8}$ consumes 6 equivalents of $\ce{O2}$, if $n_1$ of $\ce{C4H8}$ is not consumed, then $6n_1$ of $\ce{O2}$ is not consumed. Likewise, $4n_1$ of $\ce{CO2}$ and $\ce{H2O}$ were not produced.

$$\ce{C4H8 + (6-6n_1)O2 -> n_1 C4H8 + (4-4n_1)CO2 + (4-4n_1)H2O}$$

Now, I have rendered your problem down to one variable: $n_1$. Your other variables can be written as equations involving $n_1$ by comparing your chemical equation to mine (they are equivalent). You have enough equations in $n_1$ and $n_2$ to solve for $n_1$, $n_2$, and eventually $\phi$, which then leads to the values you really want.

share|improve this answer
Butene.Thanks.Could you also please take a look at my other question,I'm confused between using equilibrium constants for partial pressure and concentration. chemistry.stackexchange.com/questions/5218/… –  Sunny Marella Jul 9 '13 at 5:09
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.