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.

  2KClO_3(s) ---------- 2KCl_(s) + 3O_2(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) ----------- ZnCl_2(aq) + H_2(g) 

Solution

Zn_(S) +2HCl_(aq) ----------- ZnCl_2(aq) + H_2(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) ----------- ZnCl_2(aq) + H_2(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)

2KClO_3(s) ----------- 2KCl_(s) + 3O_2(g) 

Solution

2KClO_3(s) ----------- 2KCl_(s) + 3O_2(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

2KClO_3(s) ----------- 2KCl_(s) + 3O_2(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) ----------- FeCl_2(aq)  + H_2(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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