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)

aaronq · Answer

a mole of gas in ideal conditions is 22.414 L

you have 8.9L so divide by 22.414L, so you have 0.397 moles

in the balanced equation you have a ratio methane:water, 1:2. so you have twice as many moles, 0.397*2= 0.794 moles of H2O vapour, convert that back to Liters by multiplying by 22.414 L   and  you have 17.8 L of water vapour

aaronq · Answer

you could've just multiplied the amount of litters by 2, but i wanted to show you where it came from

anonymous · Answer

thank you!