# 0.330 mol of octane is allowed to react with 0.820 mol of oxygen. Which is the limiting reactant?

Apr 9, 2016

Dioxygen gas is the limiting reagent. We need the balanced equation representing the combustion of octane:

#### Explanation:

We need the balanced equation representing the combustion of octane:

${C}_{8} {H}_{20} \left(l\right) + 13 {O}_{2} \left(g\right) \rightarrow 8 C {O}_{2} \left(g\right) + 10 {H}_{2} O \left(g\right)$

Is this stoichiometrically balanced? Do not trust my arithmetic!

The balanced equation specifies that $1$ $\text{mole}$ of octane requires $13$ $\text{moles}$ of dioxygen gas for complete combustion. Equivalently, that each $114$ $g$ of octane requires some $336$ $g$ dioxygen for complete combustion. From where do I get these numbers?

Most of the time, in internal combustion, and diesel engines, and furnaces we would expect SOME of the hydrocarbon to combust incompletely to give $C O$, and $C$ (as soot) as oxidation products. I could represent this reaction by listing carbon and carbon monoxide as combustion products and balancing the equation appropriately:

${C}_{8} {H}_{20} \left(l\right) + \frac{23}{2} {O}_{2} \left(g\right) \rightarrow 6 C {O}_{2} \left(g\right) + C O \left(g\right) + C \left(s\right) + 10 {H}_{2} O \left(g\right)$

Would I know that I produced the $1$ mole of soot and carbon monoxide in the reaction? Of course, I would not. I would have to analyze the flue gases to make a stoichiometric appraisal. Nevertheless, this reaction represents an incomplete combustion.

How would energy transfer in these reactions? Would they produce or consume energy? Why?