How many liters of water vapor can be produced if 8.9 liters of methane gas (CH4) are combusted, if all measurements are taken a
t the same temperature and pressure? Show all of the work used to solve this problem.
CH4 (g) + 2O2 (g) yields CO2 (g) + 2H2O (g)
1 answer:
We know, mole of gas in ideal conditions is 22.414 L
Here, CH4 volume = 8.9L
So, number of moles = 8.9 / 22.414 = 0.397 moles
Here, in balanced equation we have a ratio CH4 : H2O = 1:2.
So, mole so water = , 0.397*2= 0.794 moles
In liters it would be: 0.794 * 22.414 = 17.80 L
In short, Your Answer would be 17.80 Liters
Hope this helps!
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