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Application of Gay-Lussac’s Law of Combining Volumes

Gay Lussac’s of combining volumes states that gases react in simple ratio with one another and to volumes of the products provided that temperature and pressure remain constant. In this article, you will understand how to apply this law in calculation by studying the following examples below: 1.     2H 2 + O 2 → H 2 O In the reaction above, what volume of hydrogen would be left over when 300cm 3 of oxygen and hydrogen are exploded in a sealed tube?   1cm 3 of oxygen = 2cm 3 of hydrogen 300cm 3 of oxygen = 2 x 300 = 600cm 3 Volume of left over = 1000 – 600 = 400cm 3 2.     Calculate the volume of carbon (II) oxide required to react with 40cm 3 of oxygen. 2CO + O 2 → 2CO 2 1cm 3 of oxygen = 2cm 3 of CO 40cm 3 of oxygen = 2 x 40 = 80cm 3 3.     Calculate the volume of residual gases that would be produced when 100cm 3 of sulphur (IV) oxide reacts with 20cm 3 of oxygen    2SO 2 + O 2 → 2SO 3 1cm 3 of O 2 = 2cm 3 20cm 3 of O 2 = 2 x 20 = 40cm 3
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Application of Gay-Lussac’s Law of Combining Volumes

Gay Lussac’s of combining volumes states that gases react in simple ratio with one another and to volumes of the products provided that temperature and pressure remain constant.

In this article, you will understand how to apply this law in calculation by studying the following examples below:

1.    2H2 + O2 → H2O In the reaction above, what volume of hydrogen would be left over when 300cm3 of oxygen and hydrogen are exploded in a sealed tube?

 

1cm3 of oxygen = 2cm3 of hydrogen

300cm3 of oxygen = 2 x 300 = 600cm3

Volume of left over = 1000 – 600 = 400cm3

2.    Calculate the volume of carbon (II) oxide required to react with 40cm3 of oxygen. 2CO + O2 → 2CO2

1cm3of oxygen = 2cm3 of CO

40cm3 of oxygen = 2 x 40 = 80cm3

3.    Calculate the volume of residual gases that would be produced when 100cm3 of sulphur (IV) oxide reacts with 20cm3 of oxygen    2SO2 + O2 → 2SO3

1cm3 of O2 = 2cm3

20cm3 of O2 = 2 x 20 = 40cm3

Volume of SO2 left = 100 -40 = 60cm3

The volume of SO3 produced

1cm3 of O2 = 2cm3 of SO3

20cm3 of O2 = 2 x20 = 40cm3of SO3

Total residual gases = 60 + 40 = 100cm3

4.    400cm3 of air containing 21% oxygen if 60cm3 of carbon (II) oxide burns completely with the air, calculate the volume of gases remaining.

 

Vol of oxygen in the air

21/100 x400 = 84cm3

But Vol. of oxygen that will react with CO

From the equation,

2cm3 of CO = 1cm3 of oxygen

60cm3 of CO = ½ x 60 = 30cm3

 

Vol. of remaining oxygen = 84-30 = 54cm3

Vol. of carbon (IV)oxide = 60cm3

Total residual gases = 60+54 = 114cm3

 

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