# At a given temperature, an 8*mol quantity of methane that occupies a volume of 200*L is combusted with stoichiometric oxygen gas. Can you address...?

## $\text{(i) the molar quantity of dioxygen gas required for equivalence;}$ $\text{(ii) the volume of dioxygen gas required for equivalence;}$

Oct 12, 2016

We need a stoichiometrically balanced equation:

$C {H}_{4} \left(g\right) + 2 {O}_{2} \left(g\right) \rightarrow C {O}_{2} \left(g\right) + 2 {H}_{2} O \left(l\right)$

#### Explanation:

The balanced chemical equation tells us that $\text{1 mole/1 equiv}$ of methane gas requires $\text{2 moles/2 equiv}$ of dioxygen gas for complete combustion.

We started with $4.0 \cdot m o l$ of methane, and thus we require $8.0 \cdot m o l$ of dioxygen gas to combust this completely. This represents a mass of $4.0 \cdot \cancel{m o l} \times 16.01 \cdot g \cdot \cancel{m o {l}^{-} 1} = 64.0 \cdot g$ methane, and how many grams of dioxygen gas?

Since the volume is specified, there thus must be TWICE the volume of dioxygen compared to methane for stoichiometric equivalence.