# Consider the combustion of methane: Ch4(g)+2O2(g) --> CO2(g)+2H2O(g), suppose 2.8 moles of methane are allowed to react with 3 moles of oxygen, what is the limiting reactant?

Oct 28, 2015

${O}_{2}$ is the limiting reactant.

#### Explanation:

There are different methods of finding the limiting reactant, here is the simplest to my opinion.

The reaction is:
$C {H}_{4} \left(g\right) + \textcolor{b l u e}{2} {O}_{2} \left(g\right) \to C {O}_{2} \left(g\right) + 2 {H}_{2} O \left(g\right)$

Find the molar ratio between the experimental number of moles and theoretical number of moles of the reactants:

For $C {H}_{4}$: $\frac{2.8 \cancel{m o l}}{1 \cancel{m o l}} = 2.8$.
The $1 \text{ mol}$ is taking from the coefficient of $C {H}_{4}$ from the balanced equation.

For ${O}_{2}$: $\frac{3 \cancel{m o l}}{\textcolor{b l u e}{2 \cancel{m o l}}} = 1.5$.
The $2 \text{ mol}$ is taking from the coefficient of ${O}_{2}$ from the balanced equation.

The reactant that gives the smaller molar ratio is the limiting reactant which is in this case the oxygen ${O}_{2}$

More Explanation:
To consume the $2.8 \text{ moles of } C {H}_{4}$ we need $5.6 \text{ moles of } {O}_{2}$ since the molar ratio is 1:2. We have only $3 \text{ moles of } {O}_{2}$; therefore, ${O}_{2}$ is the limiting reactant.