How do you determine the moles of the excess reactant?

Jan 2, 2018

How, well first you need the stoichiometric equation....

Explanation:

Hydrocarbon combustion is a typical reaction that occurs in internal combustion engines all over the world. The typical fuel is hexanes....${C}_{6} {H}_{14}$...for which density is typically $\rho = 0.65 \cdot g \cdot m {L}^{-} 1$.

We have say $1 \cdot L$ of hexanes to combust (carefully!), how much carbon dioxide will we generate?

And so we address the stoichiometric equation....

${C}_{6} {H}_{14} \left(l\right) + \frac{19}{2} {O}_{2} \left(g\right) \rightarrow 6 C {O}_{2} \left(g\right) + 7 {H}_{2} O \left(l\right) + \Delta$

And thus................ $\text{moles of hexanes} = \frac{1000 \cdot m L \times 0.65 \cdot g \cdot m {L}^{-} 1}{86.18 \cdot g \cdot m o {l}^{-} 1} = 7.54 \cdot m o l$.

Now under the given scenario, clearly, hexanes is the reagent in deficiency, and dioxygen is the reagent in excess..

Given the stoichiometry, I will generate approx. $45 \cdot m o l$ of carbon dioxide...and under standard conditions we could measure the volume of the gas produced. Certainly in the internal combustion engine, some of the fuel will undergo incomplete combustion to give particulate carbon, and $C O$.