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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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Stoichiometry of Chemical Reaction involving Volume of Gases part two

 


  In this second part, we are going to be looking at the stoichiometry of reaction involving the volume of gases. Just like what we saw in mass and mole, there is a relationship which also exists between mass and volume or mole and volume that is what we are going to look at.

   A close look at a balanced chemical equation shows the ratio in which the reactants react and the ratio in which the products are produced. However, an experiment has shown that at standard temperature and pressure, gases have an equal volume which is numerically equal to 22.4 dm3 or 22400cm3. This volume is known as the molar volume of gases.

1mole of any gas at STP = 22.4dm3

32g of oxygen at stp =   22.4dm3

Let us consider the decomposition of potassium trioxochlorate (V) compound.

  2KClO3(s) ---------- 2KCl(s) + 3O2(g)  

2 moles of KClO3 will decompose to give 2 moles of KCl and 3 moles of oxygen gas. Since the molar volume gas is 22.4dm3. The reacting mass of KClO3 is 2x 122.5g, and the mass of KCl produced is 2x74.5g while the volume of oxygen produced is 3x22.4dm3, according to the chemical equation. You can now see how important knowledge of chemical equations is. Let's take a look at some example questions to help us drive the topic home.

Example question 1

Calculate the volume of hydrogen produced, if 12g of zinc metal is added to excess hydrochloric acid. (H=1, Cl=35.5,Zn = 65, molar volume = 22.4dm3)

Zn(S) +2HCl(aq) ----------- ZnCl2(aq) + H2(g) 

Solution

Zn(S) +2HCl(aq) ----------- ZnCl2(aq) + H2(g) 

1mole                                               1mole

65g/mol                                           22.4dm3

From the equation,

Reacting mass of Zn = 1mole x 65g/mole = 65g

65g of Zn = 22.4dm3

12g of Zn = 22.4/65 x12/1 = 4.14dm3   or 4140cm3

By considering the number of moles involved in the reaction

Zn(S) +2HCl(aq) ----------- ZnCl2(aq) + H2(g) 

1mole                                               1mole

From the equation,

1mole of Zn = 1mole of H2

Find the number of moles in 12g of Zn

Mole = reacting mass/molar mass

Reacting mass given = 12g

Molar mass of Zn = 65g/mole

Mole = 12 /65 = 0.1846mole

Since 1mole of Zn = 1mole of H2

Then 0.1846mole of Zn = 0.1846mole

Volume of H2 in 0.1846mole =0.1846 x 22.4 =4.135dm3 or 4.14dm3 or 4140cm3

Example question 2

If 30g of potassium trioxochlorate (V) is heated in the presence of a catalyst, calculate the volume of oxygen produced at s.t.p. (Cl =35.5, O=16, K= 39, molar volume =22.4dm3)

2KClO3(s) ----------- 2KCl(s) + 3O2(g) 

Solution

2KClO3(s) ----------- 2KCl(s) + 3O2(g) 

 2mole                                 3mole

122.5g/mole                      22.4dm3

From the equation

Reacting mass of KClO3 = 2mole x 122.5g/mol =245g

Volume of O2   produced = 3x 22.4dm3 =67.2dm3

If 245g of KClO3 = 67.2dm3 of O2

Then

30g of KClO3 =67.2/245 x 30 = 8.23dm3 or 8230cm3

By considering the number of moles involved in the reaction

2KClO3(s) ----------- 2KCl(s) + 3O2(g) 

 2mole                                 3mole

From the equation,

2 moles of KClO3 = 3moles oxygen

Find the number of moles in 30g of KClO3

Mole = reacting mass / molar mass

Reacting mass given= 30g

Molar mass =122.5g/mol

Mole = 30/122.5 = 0.2449

If 2 moles of KClO3 = 3 moles of O2

 Then 0.2449 mole of KClO3 = 3/2x0.2449 = 8.23dm3 or 8230cm3

Example question 3

Calculate the mass of iron metal required to liberate 25cm3 of hydrogen, if the metal is added to excess hydrochloric acid. (Fe =56, molar volume =22.4dm3)

  Fe(s) + 2HCl(aq) ----------- FeCl2(aq)  + H2(g)

  Solution

From the equation

22400cm3 (1000x 22.4dm3) of H2 = 56g of Fe

25cm3 of H2   = 56/22400x 25/1 = 0.063g of Fe

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