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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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Mole, Molar mass and Molar Volume

                                                                                                            


 Mole

A mole of any substance is the amount of that substance which contains many elementary particles as the number of atoms in 12 grams of carbon -12. These elementary particles can be atoms, ions, electrons, molecules etc. A mole is a unit used in measuring the amount of particles.

  The number of particles in one mole of any substance is equal to 6.02 x 1023. This number is also known as Avogadro’s number.

     Molar Mass

This is the mass of one mole of any substance expressed in grams. It is also the formula mass which is the sum of all atomic masses of all atoms in a compound expressed in grams.

Example question 1

Calculate the molar mass of the following substances

i)                               sodium chloride

ii)                             chlorine

iii)                           Oxygen

(Na=23, Cl =35.5, O=16

 Solution

 

 

Sodium chloride

The relative atomic mass of Na is 23 and that of Chlorine is 35.5

The formula of sodium chloride is NaCl,

23 + 35.5 = 58.5

1 mole of sodium chloride contains 58.5g.

                Chlorine

Chlorine is a diatomic molecule, Cl2

The relative atomic mass of chlorine is 35.5

Cl2 = 2 x35.5 = 71

1 mole of Chlorine contains 71g

   Oxygen

Oxygen is also a diatomic molecule and its relative atomic mass is 16

O2= 2 x 16 =32

1mole of oxygen contains 32g

   Molar Volume

Molar volume is the volume occupied by one mole of any gas at standard temperature and pressure (s.t.p). It is equal to 22.4dm3 or 22400cm3

22.4dm3 = 6.02 x1023 = 1mole of any gas at s.t.p.

Example question 2

In a laboratory preparation of hydrogen, 5.4g of potassium reacts with water as follows:

                      2K(s) + 2H2O(l) ------ 2KOH(aq) + H2(g)

Calculate:

a)   The mass of water which reacts with 5.4g of potassium

b)   The volume of hydrogen produced at s.t.p

c)    The mass of hydroxide ions produced

(K=39, H=1, O=16, molar volume of gas =22.4dm3, Avogadro’s number = 6.02x1023)

            Solution

2K + 2H2O --------- 2KOH +  H2

From the equation,

a.         2moles of K = 2moles of H2O

             2 x 39g of K = 2 x18g of H2O

            78g of K = 36g of H2O

           5.4g of K = 36/78 x 5.4

            = 2.49g

b.     2 moles of K = 1 mole of H2

78g of K = 22.4dm3 or 22400cm3

5.4g of K = 22400/78 x 5.4

= 1550 cm3

 

c.     2K + 2H2O ------- K+ + 2OH-  + H2

2 moles of K = 2moles of OH- ions

78g of K = 2 x 6.02x1023

5.4g of K = 1.204x1023/78x5.4

= 8.34x1022 of OH-

 

 

 

 

 

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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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