# If we combust 23.6*g of methanol, how much water will be produced?

Mar 22, 2017

I get $26.5 \cdot g$ of water from the combustion.......

#### Explanation:

And we need a stoichiometric equation that represents the complete combustion of methanol:

${H}_{3} C O H \left(l\right) + \frac{3}{2} {O}_{2} \left(g\right) \rightarrow C {O}_{2} \left(g\right) + 2 {H}_{2} O \left(l\right)$

Is it balanced? It looks like it to me.

And so we combust $23.6 \cdot g$ of methanol, which represents a molar quantity of:

$\frac{23.6 \cdot g}{32.04 \cdot g \cdot m o {l}^{-} 1} = 0.737 \cdot m o l$

Given the stoichiometry, at most, we can get $2 \times 0.737 \cdot m o l = 1.47 \cdot m o l$ water. And this represents a mass of $1.47 \cdot m o l \times 18.01 \cdot g \cdot m o {l}^{-} 1 = 26.5 \cdot g$, which was a book option. Claro?

If you are unsatisfied, I am willing to try again.