# Gay-Lussac’s Law of Combining Volumes

 Science > Chemistry >You are Here • Joseph Louis Gay Lussac ( 6 December 1778 – 9 May 1850) was a French chemist and physicist.
• He is known mostly for two laws related to gases, and for his work on alcohol-water mixtures

### Gay-Lussac’s Law of Combining Volumes of Gases:

• A French chemist Joseph L. Gay – Lussac in 1809, put forward this law.

#### Statement :

• Whenever gases take part in a chemical reaction, either as reactants or as products, they do so in simple proportions by Volumes. Provided the volumes of gases are measured at the same temperature and pressure,

#### Illustration 1:

• Consider following reaction

H2(g)      +     Cl2(g)     →     2HCl
1vol               1vol                   2 vol
Thus the simple ratio of volumes is 1 : 1 : 2

#### Illustration 2 :

• Consider following reaction

N2(g)      +    3 H2(g)     →     2NH3(g)
1vol            3vol                   2 vol
Thus the simple ratio of volumes is 1 : 3 : 2

### Applications of Gay-Lussac’s Law of Combining Volumes of Gases:

#### Example – 1:

• Calculate the volume of oxygen required for the complete combustion of 0.25 dm3 of methane at STP.
• Solution:

CH4(g)   +   2O2(g)    →   CO2(g)    +   2H2O(g)
1 vol            2 vol           1 vol                  2 vol

By Gay-Lussac’s law of combining volumes of gases

1 vol of methane requires 2 vol of oxygen for complete combustion.
Hence  0.25 dm3 of methane requires 2 x 0.25 = 0.50 dm3 of oxygen.

#### Example – 2:

• Calculate the volume of hydrogen required for the complete hydrogenation of 0.25 dm3 of ethyne at STP.
• Solution:

C2H2(g)    +    2H2(g)  →     C2H6(g)
1 vol           2vol                1 vol

By Gay-Lussac’s law of combining volumes of gases

1 vol of ethyne requires 2 vol of hydrogen for complete hydrogenation.
Thus 0.25 dm3 of ethyne requires 2 x 0.25 = 0.50 dm3 of hydrogen for complete hydrogenation.

#### Example – 3:

• Calculate the volume of hydrogen required for the complete hydrogenation of 0.25 dm3 of ethylene at STP.
• Solution:

C2H4(g)    +    H2(g)  →     C2H6(g)
1 vol           1vol                1 vol

By Gay-Lussac’s law of combining volumes of gases

1 vol of ethylene requires 1 vol of hydrogen for complete hydrogenation.

Thus 0.25 dm3 of ethylene requires 1 x 0.25 = 0.25 dm3 of hydrogen for complete hydrogenation.

#### Example – 4:

• Calculate the volume of oxygen required for the complete combustion of 0.25 mol of methane at STP.
• Solution:

CH4(g)   +   2O2(g)    →   CO2(g)    +   2H2O(g)
1 mol            2 mol           1 mol                2 mol

By Gay-Lussac’s law of combining volumes of gases

1 mol of methane requires 2 mol of oxygen for complete combustion.

Thus 0.25 mol of methane requires 2 x 0.25 = 0.50 mol of oxygen.
One mole of any gas occupies 22.4 dm3 by volume at STP.
Volume of oxygen required = 22.4 x No. of moles = 22.4 x 0.5 = 11.2 dm3

#### Example – 5:

• Calculate the volume of oxygen required for the complete combustion of 0.5 dm3 of H2S at STP.
• Solution:

2H2S(g)   +   3O2(g)    →   2SO2(g)    +   2H2O(g)
2 vol           3 vol           2 vol                  2 vol

By Gay-Lussac’s law of combining volumes of gases

2 vol of H2S requires 3 vol of oxygen for complete combustion.
1 vol of H2S requires 3/2 = 1.5 vol of oxygen for complete combustion.
Thus 0.5 dm3 of H2S requires 1.5 x 0.5 = 0.75 dm3 of oxygen for complete combustion.

#### Example – 6:

• Calculate the volume of oxygen required at STP for the complete combustion of 5.0 dm3 of ethane at 295 K and 0.993 x 105 Nm-2
• Solution:

2C2H6(g)   +   7O2(g)    →   4CO2(g)    +   6H2O(g)
2 vol            7 vol              4 vol                6 vol

By Gay-Lussac’s law of combining volumes of gases

2 vol of C2H6 requires 7 vol of oxygen for complete combustion.
1 vol of C2H6 requires 7/2 = 3.5 vol of oxygen for complete combustion.
Thus 5 dm3 of C2H6 requires 3.5 x 5 = 17.5 dm3 of oxygen for complete combustion.
P = 0.993 x 105 Nm-2, T = 295 K, V = 17.5 dm3
Po = 1.013 x 105 Nm-2, To = 273 K, Vo = ? Thus the volume of oxygen required at STP is 15.88 dm3.

#### Example – 7:

• 6.0 dm3 of hydrogen is reacted with 2.4 dm3 of oxygen in a closed chamber. Calculate composition of resulting mixture.
• Solution:

2H2(g)    +    O2(g)  →    2 H2O(g)
2 vol           1 vol               2 vol

By Gay-Lussac’s law of combining volumes of gases
The ratio of the volume of hydrogen to that of oxygen is 2 : 1.
In this case oxygen is limiting reagent and hydrogen is excess reagent.
2.4 dm3 of oxygen can combine with 2 x 2.4 = 4.8 dm3 of hydrogen to form 2 x 2.4 = 4.8 dm3 of water vapours.
Thus unreacted hydrogen = 6.0 – 4.8 = 1. 2  dm3.
Thus resulting mixture contains 4.8 dm3 of water vapours and 1.2 dm3 of unreacted hydrogen.

#### Example – 8:

• 15 litres of nitrogen is made to react with 30 litres of hydrogen to prepare ammonia. Calculate composition of resulting mixture.
• Solution:

N2(g)      +    3 H2(g)     →     2NH3(g)
1vol            3vol                   2 vol

By Gay-Lussac’s law of combining volumes of gases

The ratio of the volume of nitrogen to that of hydrogen is 1 : 3.
In this case, hydrogen is limiting reagent and nitrogen is excess reagent.
3 x 10 = 30 litres of hydrogen can combine with 1 x 10 = 10 litres of nitrogen to form 2 x 10 = 20 litres of ammonia.
Thus unreacted nitrogen = 15.0 – 10.0 = 5 litres.
Thus resulting mixture contains 20 litres of ammonia and 5 litres of unreacted nitrogen.

#### Example – 9:

• 200 dm3 of hydrogen gas is allowed to react with 250 dm3 of chlorine gas. Calculate composition of resulting mixture.
• Solution:

H2(g)      +     Cl2(g)     →     2HCl
1vol               1vol                   2 vol

By Gay-Lussac’s law of combining volumes of gases
The ratio of the volume of hydrogen to that of chlorine is 1 : 1.
In this case, hydrogen is limiting reagent and chlorine is excess reagent.
200 dm3 of hydrogen can combine with 200 dm3 of chlorine to form 2 x 200 =400 dm3of hydrogen chloride.
Thus unreacted chlorine = 250 – 200 = 50 dm3.
Thus resulting mixture contains 400 dm3 of hydrogen chloride and 50 dm3 of unreacted chlorine.

#### Example – 10:

• 10 dm3 of hydrogen gas is allowed to react with 15 dm3 of chlorine gas. Calculate composition of resulting mixture.
• Solution:

H2(g)      +     Cl2(g)     →     2HCl
1vol               1vol                   2 vol

By Gay-Lussac’s law of combining volumes of gases
The ratio of the volume of hydrogen to that of chlorine is 1 : 1
In this case, hydrogen is limiting reagent and chlorine is excess reagent.
10 dm3 of hydrogen can combine with 10 dm3 of chlorine to form 2 x 10 = 20 dm3 of hydrogen chloride.
Thus unreacted chlorine = 15 – 10 = 5 dm3.
Thus resulting mixture contains 20 dm3 of hydrogen chloride and 5 dm3 of unreacted chlorine

 Science > Chemistry >You are Here

### One Comment

1. Md Afzal ali

very well explained. extremely impressed with examples. thank you so much.