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

1 Answer
Mar 22, 2017

Answer:

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.