# When octane is completely combusted, what is the whole number ratio between the dioxygen reactant, and the product hydrogens?

Feb 22, 2017

Well, clearly it's $25 : 18$.
${C}_{8} {H}_{18} \left(l\right) + \frac{25}{2} {O}_{2} \left(g\right) \rightarrow 8 C {O}_{2} \left(g\right) + 9 {H}_{2} O \left(l\right)$
And the stoichiometry precisely specifies that $\text{25 equiv}$ of oxygen gas are required for complete combustion of $\text{2 equiv}$ of $\text{octane}$, to give $\text{16 equiv}$ of $\text{carbon dioxide}$ and $\text{18 equiv}$ of $\text{water.}$