Chapter 9.2, Problem 2RQ

Interpretation Introduction

Interpretation: The mass of ${\text{KClO}}_3$ required and mass of $\text{KCl}$ producedfrom 15.0 g of ${\text{O}}_2$ needs to be calculated for the given reaction.

Concept introduction: The ratio of mass of substance to its molar mass is said to be number of moles of that substance.

Answer to Problem 2RQ

The mass of ${\text{KClO}}_3$ required is 38.36 g and mass of $\text{KCl}$ produced is 23.33 g 15.0 g of ${\text{O}}_2$ for the given reaction.

Explanation of Solution

Given:

The unbalanced reaction:

  ${\text{KClO}}_3(s)\to \text{KCl}(s)+{\text{ O}}_2(g)$

The reaction can be balanced by writing coefficient 2 before ${\text{KClO}}_3$ in the reactant side and 2 and 3 before $\text{KCl}$ and ${\text{O}}_2$ in the product side. Thus, the balanced reaction is:

  ${\text{2KClO}}_3(s)\to 2\text{KCl}(s)+{\text{ 3O}}_2(g)$

From the balanced chemical reaction, the mole ratio of ${\text{KClO}}_3$ and ${\text{O}}_2$ is 2:3 and mole ratio of $\text{KCl}$ and ${\text{O}}_2$ is 2:3.

The mole ratio for the given condition is calculated as:

The molar mass of oxygen 31.998 g/mol. The number of moles of oxygen is calculated as shown:

  $\begin{array}{l}\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}\\ \text{Number of moles}=\frac{\text{15 g}}{\text{31}\text{.998 g/mol}}\\ \text{Number of moles}=0.47\text{ mol}\end{array}$

So,

• The number of moles of ${\text{KClO}}_3$ for 0.47 mol of ${\text{O}}_2$ is:

  $\text{0}{\text{.47 mol of O}}_{\text{2}}\text{×}\frac{{\text{2 mol of KClO}}_3}{{\text{3 mol of O}}_{\text{2}}}\text{ = 0}{\text{.313 mol of KClO}}_3$

Now, the mass of ${\text{KClO}}_3$ required is calculated as:

The molar mass of ${\text{KClO}}_3$ is 122.55 g/mol. So,

  $\begin{array}{l}\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}\\ \text{0}\text{.313 mol}=\frac{\text{Mass}}{\text{122}\text{.55 g/mol}}\\ \text{Mass = 0}\text{.313 mol}\times \text{122}\text{.55 g/mol}\\ \text{Mass = }38.36\text{ g}\end{array}$

• The number of moles of $\text{KCl}$ for 0.47 mol of ${\text{O}}_2$ is:

  $\text{0}{\text{.47 mol of O}}_{\text{2}}\text{×}\frac{\text{2 mol of KCl}}{{\text{3 mol of O}}_{\text{2}}}\text{ = 0}\text{.313 mol of KCl}$

Now, the mass of $\text{KCl}$ required is calculated as:

The molar mass of $\text{KCl}$ is 74.55 g/mol. So,

  $\begin{array}{l}\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}\\ \text{0}\text{.313 mol}=\frac{\text{Mass}}{\text{74}\text{.55 g/mol}}\\ \text{Mass = 0}\text{.313 mol}\times \text{74}\text{.55 g/mol}\\ \text{Mass = 23}\text{.33 g}\end{array}$

Hence, the mass of ${\text{KClO}}_3$ required is 38.36 g and mass of $\text{KCl}$ produced is 23.33 g 15.0 g of ${\text{O}}_2$ for the given reaction.