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I am trying to balance out the equation for combustion of octane in $120\,\%$ of excess air with products $\ce{CO2},$ $\ce{H2O},$ $\ce{N2},$ and $\ce{O2}$:

$$\ce{C8H18 + 1.2$a$(O2 + 3.76 N2) -> $b$ CO_2 + $c$ H2O + $d$ O2 + $e$ N2}$$

My approach was as follows:

$$ \begin{align} &\ce{C}: & b &= 8 \\ &\ce{H}: & 2c &= 18 \\ &\ce{O}: & 2b + c + 2d &= 2\times 1.2a \\ &\ce{N}: & e &= 3.76\times 1.2a \end{align} $$

However, I think somewhere in my steps there's something I am not doing correctly as I am unable to determine $a$ with the system of equations I tried developing.

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    $\begingroup$ Octane combusted in air yields carbon dioxide plus water. The nitrogen in air does nothing, unless you are in, e.g., an internal combustion engine. So ignore nitrogen. $\endgroup$
    – Ed V
    Commented Jul 18, 2021 at 15:37
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    $\begingroup$ This is for an internal combustion engine. $\endgroup$
    – RMS
    Commented Jul 18, 2021 at 16:07
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    $\begingroup$ I have found the already balanced equation online, it includes all those reactants and all those products, I am trying how to figure out the balancing aspect. $\endgroup$
    – RMS
    Commented Jul 18, 2021 at 16:16
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    $\begingroup$ A chemical equation must not mention substances that are not transformed, like $\ce{N2}$and part of $\ce{O2}$ here. $\endgroup$
    – Maurice
    Commented Jul 18, 2021 at 16:24
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    $\begingroup$ Maurice I understand what you are trying to explain; however, these reaction is written including O2 and N2 and the balancing process was done including them. If you google the reaction you can see what I am talking about. $\endgroup$
    – RMS
    Commented Jul 18, 2021 at 16:26

2 Answers 2

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Since the reaction is done in excess of air, you can assume complete combustion and accordingly you should break down the reaction to only this.

Apparently the question demands an answer in the following form: $$\ce{C8H18 + 1.2$a$ (O2 + 3.76 N2) -> b CO2 + c H2O + d O2 + e N2},$$ however, this is the same as $$\begin{multline} \ce{C8H18 + x O2 + 0.2x O2 + (1.2$x$\times3.76) N2}\\ \ce{-> y CO2 + z H2O + 0.2x O2 + (1.2$x$\times3.76) N2}, \end{multline}$$ which is the same as $$\ce{C8H18 + x O2 -> y CO2 + z H2O},$$ just more complicated.

From there you can form \begin{align} \ce{C}: && y &= 8\\ \ce{H}: && z &= \frac{1}{2}\times18 = 9\\ \ce{O}: && 2 x &= 2y + z \Longleftrightarrow x = 12.5. \end{align}

Now you can obviously go back to the requested form: $$\ce{C8H18 + 15 (O2 + 3.76 N2) -> 8 CO2 + 9 H2O + 2.5 O2 + 56.4 N2}.$$

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  • $\begingroup$ (+1) And, of course, no internal combustion engine was involved, even putatively: no nitrogen oxides were produced. As well, the problem did not involve considerations of thermodynamics, e.g., the heating up of the nitrogen in the air. $\endgroup$
    – Ed V
    Commented Jul 18, 2021 at 17:49
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Key things I did not consider properly:

120% of excess air, which means 0.2 of oxygen goes free, this changes my equation to:

$$\ce{C8H18 + 1.2$a$ (O2 + 3.76 N2) -> b CO2 +c H2O + 0.2a O2 + e N2} $$

So,

\begin{align} \ce{C}:&& b &= 8\\ \ce{H}:&& 2c &= 18\\ \ce{O}:&& 2b + c + 2(0.2) &= 2\times1.2a \implies a = 12.5\\ \ce{N}:&& e &= 3.76\times1.2a = 56.4 \end{align}

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    $\begingroup$ This is exactly the same as $\ce{C8H18 + x O2 -> y CO2 + z H2O}$, just more complicated. $\endgroup$ Commented Jul 18, 2021 at 17:14
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    $\begingroup$ The point was for me to understand how to workout the problem using this kind of forms. I still need to write down the balanced from of the more complicated expression. $\endgroup$
    – RMS
    Commented Jul 18, 2021 at 18:29

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