How many liters of water vapor can be produced if 8.9 liters of methane gas (CH4) are combusted, if all measurements are taken at the same temperature and pressure?

Question

shoba · Answer

CH4 + 2O2 --- CO2 + 2H2O



I would calculate so: 
pv = nRT 
n = pv/(RT) 
n = 1,2*1,0133*10^5 Pa*0,0158 m^3/ (8,314 J/(mol*k) *300° K) 
n = 7,702*10^-1 moles 
n = 0,7702 moles. 
If 1 mole CH4 = 16g, then 0,77 moles are 0,77*16 g = 12,32 g 
------ 
The equation says: 1 mole of CH4 produces 2 moles H2O 
Now you do not know, how many moles you have, because you do not know Temperature and pressure. But at any temperature or pressure is the volume of gases (CH4 and H2O) 
proportional to volume. 
So you can calculate directly with volumes without going into moles. 
And this means: 8,9 L CH4 will produce twice as many Liters of H2O = 17,8 L.

anonymous · Answer

CH4 + 2 O2 → CO2 + 2 H2O 

(8.9 L CH4) x (2 mol H2O / 1 mol CH4) = 17.8 L H2O vapor