Question #18a97

1 Answer
Nov 6, 2017

"15.6 g CH"_3"OH"

Explanation:

Assuming that the reaction in the question is a complete combustion reaction,

CH_3OH + 3/2O_2 to CO_2+2H_2O

Molar Mass of H_2O is "18 02 g/mol"
Molar Mass of CH_3OH = "32.04 g/mol"

"17.5 g H"_2"O" *("1 mol H"_2"O")/("18.02 g H"_2"O")*("1 mol CH"_3"OH")/("2 mol H"_2"O")*("32.04 g CH"_3"OH")/("1 mol CH"_3O"H")

= "15.6 g CH"_3"OH"