# Can you represent the combustion of pentane by means of a stoichiometric equation?

Feb 21, 2017

${C}_{5} {H}_{12} \left(l\right) + 8 {O}_{2} \left(g\right) \rightarrow 5 C {O}_{2} \left(g\right) + 6 {H}_{2} O \left(l\right)$
The given equation is stoichiometrically balanced: garbage in equals garbage out. If the pentane is in stoichiometric proportion with the dioxygen and $\text{10 mol dioxygen}$ were used, then there were $\frac{5}{4} \cdot m o l$ of $\text{pentane}$ precisely, and $\frac{25}{4} \cdot m o l$ of carbon dioxide gas would result.
$\text{1 mole}$ or $\text{1 equiv}$ $\text{pentane}$ react with $\text{8 equiv}$ dioxygen to give $\text{5 equiv}$ carbon dioxide, and $\text{6 equiv}$ water. In an internal combustion engine, however, combustion is rarely so complete. The products of incomplete combustion, $C O$, and $C$, are known to occur.