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How many liters of water vapor can be produced if 13.3 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 :

Edufirst
At the same T and P, the number of mols are proportional to the volume

The stoichiometry shows that 1 mol of CH4 yields 2 mols of water vapor

Then 13.3 lites yields 2 x 13.3 liters = 26.6 liters

Answer: The volume of water vapor produced is 26.6 L.

Explanation:

To calculate the volume of water vapor when number of moles changes , we use Avogadro's Law. This law states that the volume is directly proportional to the number of moles of gas if temperature and pressure remains constant.

Mathematically,

[tex]V\propto n\\\\\frac{V_1}{n_1}=\frac{V_2}{n_2}[/tex]

where,  

[tex]V_1\text{ and }n_1[/tex] = Initial volume and number of moles

[tex]V_2\text{ and }n_2[/tex] = Final volume and number of moles

For the given chemical reaction:

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

By Stoichiometry,

1 mole of methane reacts with 2 moles of oxygen gas to produce 1 mole oc carbon dioxide and 2 moles of water vapor.

We are given:  

[tex]V_1=13.3L\\n_1=1mole\\V_2=?L\\n_2=2mole[/tex]

Putting values in above equation, we get:

[tex]\frac{13.3L}{1mol}=\frac{V_2}{2mol}[/tex]

[tex]V_2=26.6L[/tex]

Hence, the volume of water vapor produced is 26.6 L.

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