Chapter 12.2, Problem 15SP

Interpretation Introduction

Interpretation: The number of molecules of oxygen produced by the decomposition of 6.54 g of potassium chlorate ${\text{KClO}}_{\text{3}}$ needs to be calculated.

Concept Introduction: The relation between the number of moles and the mass of a substance is as follows:

  $n=\frac{m}{M}$

Here, m is mass and M is molar mass.

In 1 mol of a substance, there are $6.022\times {10}^{23}$ units of that substance.

Explanation of Solution

The given reaction is as follows:

  ${\text{2KClO}}_{\text{3}}\left(s\right)\to 2\text{KCl}\left(s\right)+3{\text{O}}_{\text{2}}\left(g\right)$

The mass of potassium chlorate ${\text{KClO}}_{\text{3}}$ is 6.54 g.

The number of moles of ${\text{KClO}}_{\text{3}}$ can be calculated as follows:

  $n=\frac{m}{M}$

Molar mass of ${\text{KClO}}_{\text{3}}$ is 122.55 g/mol. Put the values to the calculated number of moles.

  $\begin{array}{c}n=\frac{6.54\text{ g}}{122.55\text{ g/mol}}\\ =0.0534\text{ mol}\end{array}$

Now, from the given reaction, 2 mol of ${\text{KClO}}_{\text{3}}$ gives 3 mol of ${\text{O}}_{\text{2}}$ ; thus, 1 mol of ${\text{KClO}}_{\text{3}}$ will give $\frac{3}{2}$ mol of ${\text{O}}_{\text{2}}$ .

Now, 1 mol of oxygen contains $6.022\times {10}^{23}$ oxygen molecules. Thus, the number of oxygen molecules in $\frac{3}{2}$ mol of ${\text{O}}_{\text{2}}$ will be:

  $\begin{array}{c}N=\frac{\left(6.022\times {10}^{23}{\text{ O}}_2\text{ molecules}\right)\left(\frac{3\text{ mol}}{2}\right)}{1\text{ mol}}\\ =9.033\times {10}^{23}{\text{ O}}_2\text{ molecules}\end{array}$

Therefore, the number of oxygen molecules will be $9.033\times {10}^{23}{\text{ O}}_2\text{ molecules}$ .