# What volume of water results from complete combustion of a 40*L volume of methane under standard conditions?

May 5, 2017

The problem with this question is that we must ASSUME standard conditions........

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

#### Explanation:

Methane is a room temperature gas....,$\text{boiling point } =$ $- 164$ ""^@C. And of course water is a room temperature liquid.....

We assume that the methane gas is under standard conditions, i.e. $298 \cdot K$, and $1 \cdot a t m$.

And this represents $\frac{40 \cdot L}{24.5 \cdot L \cdot m o {l}^{-} 1} = 1.63 \cdot m o l$ of the hydrocarbon.

Given the stoichiometry, we get................

$\frac{2 \times 1.63 \cdot m o l \times 18.01 \cdot g \cdot m o {l}^{-} 1}{1.0 \cdot g \cdot m {L}^{-} 1}$

$\cong 60 \cdot m L$