anonymous
  • anonymous
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)
Chemistry
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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
  • anonymous
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

Looking for something else?

Not the answer you are looking for? Search for more explanations.