Chapter 18, Problem 121A

(a)

Interpretation Introduction

Interpretation: The expression for a balanced chemical equation for the reaction is to be given.

Concept Introduction: A chemical equation in which the number of atoms on both reactant as well as product side is equal is termed as a balanced chemical equation.

Answer to Problem 121A

The balanced chemical equation for the given reaction is represented as:

  ${\text{2KClO}}_{\text{3}}\to 2\text{KCl}+3{\text{O}}_{\text{2}}$

Explanation of Solution

The chemical equation for the reaction of potassium chlorate to produce potassium chloride and oxygen gas is given as:

  ${\text{KClO}}_{\text{3}}\to \text{KCl}+{\text{O}}_{\text{2}}$

Thus, the balanced chemical equation for the above reaction is given as:

  ${\text{2KClO}}_{\text{3}}\to 2\text{KCl}+3{\text{O}}_{\text{2}}$

(b)

Interpretation Introduction

Interpretation: The grams of oxygen formed by the decomposition of $4.88\text{g}{\text{KClO}}_{\text{3}}$ is to be determined.

Concept Introduction: A chemical equation in which the number of atoms on both reactant as well as product side is equal is termed a balanced chemical equation.

Answer to Problem 121A

1.91 grams of oxygen are formed by the decomposition of $4.88\text{g}{\text{KClO}}_{\text{3}}$ .

Explanation of Solution

According to the reaction,

  $\begin{array}{c}\text{Grams}\text{of}\text{oxygen}\text{formed}=\text{4}\text{.88}\text{g}{\text{KClO}}_{\text{3}}\times \frac{\text{1}\text{mol}{\text{KClO}}_{\text{3}}}{\text{122}\text{.55}\text{g}{\text{KClO}}_{\text{3}}}\times \frac{\text{3}\text{mol}{\text{O}}_{\text{2}}}{\text{2}\text{mol}{\text{KClO}}_{\text{3}}}\times \frac{\text{31}\text{.998}\text{g}{\text{O}}_{\text{2}}}{\text{1}\text{mol}{\text{O}}_{\text{2}}}\\ =1.91g\end{array}$