Question

Question #18a97

Answer 1

`"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"`