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

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!

Answer Link

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

Respuesta :

Otras preguntas

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