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

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

Is it balanced? It looks like it to me.

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

To find out how much water will be produced when 23.6 g of methanol (CH₃OH) is combusted, we need to use stoichiometry and the balanced chemical equation for the combustion of methanol:

2 CH₃OH(l) + 3 O₂(g) → 2 CO₂(g) + 4 H₂O(g)

From the equation, we see that for every 2 moles of methanol combusted, 4 moles of water are produced.

1 mole of CH₃OH has a molar mass of approximately 32.04 g.

First, we calculate the number of moles of methanol:

$23.6 \text{ g} \times \frac{1 \text{ mol}}{32.04 \text{ g/mol}} = 0.736 \text{ mol}$

Now, we can use stoichiometry to find the moles of water produced:

$0.736 \text{ mol CH₃OH} \times \frac{4 \text{ mol H₂O}}{2 \text{ mol CH₃OH}} = 1.472 \text{ mol H₂O}$

Finally, we convert the moles of water to grams:

$1.472 \text{ mol H₂O} \times \frac{18.02 \text{ g/mol}}{1 \text{ mol H₂O}} = 26.53 \text{ g}$

So, when 23.6 g of methanol is combusted, approximately 26.53 g of water will be produced.