# 0.200 mol of octane is allowed to react with 0.690 mol of oxygen, which is the limiting reactant? 2 C8H18 + 25 O2 ----> 16 CO2 + 18 H2O.

Jul 4, 2018

Well let us rebalance the equation to make our ideas of stoichiometric equivalence a bit more straightforward...

#### Explanation:

We combust octane, and we represent complete combustion in the usual way: balance the carbons as carbon dioxide; balance the hydrogens as water; and THEN balance the oxygens...

${C}_{8} {H}_{18} \left(l\right) + \frac{25}{2} {O}_{2} \left(g\right) \rightarrow 8 C {O}_{2} \left(g\right) + 9 {H}_{2} O \left(l\right)$

$\text{Moles of octane} = 0.200 \cdot m o l$; and this requires $12 \cdot \frac{1}{2} \cdot \text{equiv}$ dioxygen...i.e. $2.50 \cdot m o l$.... In the reaction dioxygen is thus the LIMITING REAGENT...

And here CLEARLY, given that dioxygen is specified to be in $0.690 \cdot m o l$ quantity....octane is in stoichiometric EXCESS...and the reaction would not proceed as written. We COULD work out precisely how much carbon dioxide would result in the case of complete combustion. More realistically, we should be aware that INCOMPLETE combustion to $C O \left(g\right)$ and $C \left(s\right)$ would occur, as it certainly does in the internal combustion, and diesel engines. And as a representation we could write...

${C}_{8} {H}_{18} \left(l\right) + 11 {O}_{2} \left(g\right) \rightarrow 6 C {O}_{2} \left(g\right) + C O \left(g\right) + C \left(s\right) + 9 {H}_{2} O \left(l\right)$

For this PROPOSED stoichiometry...dioxygen is AGAIN the limiting reagent...

To conclude we HAVE NO HANDLE on the stoichiometry of this reaction....and the question is improperly proposed...

Jul 4, 2018

It should be oxygen.

#### Explanation:

We first look at the equation:

$2 {C}_{8} {H}_{18} \left(l\right) + 25 {O}_{2} \left(g\right) \to 16 C {O}_{2} \left(g\right) \uparrow + 18 {H}_{2} O \left(l\right) + \Delta$

And so, we clearly see that $2$ moles of octane require $25$ moles of oxygen gas to burn. Since we have $0.2$ moles of octane, we require:

0.2color(red)cancelcolor(black)("mol" \ C_8H_18)*(25 \ "mol" \ O_2)/(2color(red)cancelcolor(black)("mol" \ C_8H_18))=2.5 \ "mol" \ O_2

But since we only have $0.69 \setminus \text{mol} \setminus {O}_{2}$, we clearly see that it is the limiting reactant in this case.