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[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

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3.0 liters is the answer

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

Explanation:

At STP:

1 mole of a gas occupies 22.4 liters of volume.

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

[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]

By stoichiometry of the reaction:

If [tex](1\times 22.4)=22.4L[/tex] of methane gas produces [tex](2\times 22.4)=44.8L[/tex] of water vapor.

Then, 1.5 L of methane gas will produce = [tex]\frac{44.8L}{22.4}\times 1.5L=3L[/tex] of water vapor.

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

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