Question

[05.04]when 1.5 l of methane gas combust in an excess of oxygen at standard temperature and pressure, how many liters of water vapor will be produced? ch4 (g) + 2o2 (g) yields co2 (g) + 2h2o (g) 0.75 l 1.5 l 3.0 l 6.0 l

Answer 1

3.0 liters is the answer

Answer 2

Answer: The volume of water vapor produced will be 3 L.

Step-by-step explanation:

At STP:

1 mole of a gas occupies 22.4 liters of volume.

For the reaction of combustion of methane, the equation follows:

`CH_4+2O_2\rightarrow CO_2+2H_2O`

By stoichiometry of the reaction:

If
`(1* 22.4)=22.4L` of methane gas produces
`(2* 22.4)=44.8L` of water vapor.

Then, 1.5 L of methane gas will produce =
`(44.8L)/(22.4)* 1.5L=3L` of water vapor.

Hence, the volume of water vapor produced will be 3 L.