Question

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

Answer 1

I get `26.5*g` of water from the combustion.......

Explanation:

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

`H_3COH(l) + 3/2O_2(g) rarr CO_2(g) + 2H_2O(l)`

Is it balanced? It looks like it to me.

And so we combust `23.6*g` of methanol, which represents a molar quantity of:

`(23.6*g)/(32.04*g*mol^-1)=0.737*mol`

Given the stoichiometry, at most, we can get `2xx0.737*mol=1.47*mol` water. And this represents a mass of `1.47*molxx18.01*g*mol^-1=26.5*g`, which was a book option. Claro?

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