What volume of water results from complete combustion of a #40*L# volume of methane under standard conditions?

1 Answer
May 5, 2017

Answer:

The problem with this question is that we must ASSUME standard conditions........

#CH_4(g) + 2O_2(g) rarr CO_2(g) +2H_2O(l)#

Explanation:

Methane is a room temperature gas....,#"boiling point "=# #-164# #""^@C#. And of course water is a room temperature liquid.....

We assume that the methane gas is under standard conditions, i.e. #298*K#, and #1*atm#.

And this represents #(40*L)/(24.5*L*mol^-1)=1.63*mol# of the hydrocarbon.

Given the stoichiometry, we get................

#(2xx1.63*molxx18.01*g*mol^-1)/(1.0*g*mL^-1)#

#~=60*mL#