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? Show all of the work used to solve this problem.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Question

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? Show all of the work used to solve this problem. 

CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g)

anonymous · Answer

at same temperature and pressure, volume is directly proportional to the number of moles.
so,
$$\frac{ volume of methane combusted }{ volume of water vapour produced } = \frac{ number of moles of methane combusted }{ number of moles of water vapour produced. }$$
and volume of methane = 8.9L
number of moles of methane reacted = 1
number of moles of water vapour produced = 2

putting this data in the above equation,
we get,

volume of water vapour produced = 17.8L.

anonymous · Answer

No.. This is not a practice exam. This is a module quiz from FLVS and it's cheating if you use this answer. Plus it has a obvious mistake in it. Mr. Carlyle 
FLVS Chemistry Instructor 

This question is a violation of the OpenStudy Guidelines. The question is from an online school plus a violation by solicitation for members to unknowingly assist them cheating on an exam. 30653065chem30653065