Chapter 9, Problem 167QP

(a)

Interpretation Introduction

Interpretation:

Whether $11.25\text{ mol}$ of oxygen is expected or not is to be determined.

Explanation of Solution

Oxygen is produced by the following reaction:

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

According to the reaction,

$\begin{array}{l}2{\text{ mol KClO}}_{\text{3}}\text{ produces}=3{\text{ mol O}}_{\text{2}}\\ 1{\text{ mol KClO}}_{\text{3}}\text{ produces}=\frac{3}{2}{\text{mol O}}_{\text{2}}\end{array}$

The molar mass of ${\text{KClO}}_{\text{3}}$ is $122.6{\text{ g mol}}^{-1}$ . So, $1\text{ mol}$ of ${\text{KClO}}_{\text{3}}$ contains $122.6\text{ g}$ ${\text{KClO}}_{\text{3}}$ .

$\begin{array}{c}122.6{\text{ g KClO}}_{\text{3}}\text{ produces}=\frac{3}{2}{\text{mol O}}_{\text{2}}\\ {\text{1 g KClO}}_{\text{3}}\text{ produces}=\frac{3}{2\times 122.6}{\text{mol O}}_{\text{2}}\end{array}$

When $7.2\text{g}$ ${\text{KClO}}_{\text{3}}$ completely reacts and the oxygen gas is collected over water at $29°\text{C}$ and $765\text{ torr}$ , the moles of oxygen expected to be:

$\begin{array}{c}7.5{\text{ g KClO}}_{\text{3}}\text{ produces}=\frac{3}{2\times 122.6}\times \text{7}{\text{.5 mol O}}_{\text{2}}\\ =0.09176{\text{ mol O}}_{\text{2}}\end{array}$

So, $11.25\text{ mol}$ of oxygen is not expected and Statement A is incorrect.

(b)

Interpretation Introduction

Interpretation:

Whether the partial pressure of water at at $29°\text{C}$ is $765\text{ torr}$ or not is to be determined.

Explanation of Solution

When $7.2\text{g}$ ${\text{KClO}}_{\text{3}}$ completely reacts and the oxygen gas is collected over water at $29°\text{C}$ and $765\text{ torr}$ . The total pressure of the reaction that is the sum of the partial pressure of water and oxygen is $765\text{ torr}$ . The partial pressure of the water is determined by substituting data from partial pressure table 9.2, that is, $30\text{ torr}$ at $29°\text{C}$ .

So, the partial pressure of water at $29°\text{C}$ is $30\text{ torr}$ and statement B is incorrect.

(c)

Interpretation Introduction

Interpretation:

Whether the partial pressure of oxygen is $735\text{ torr}$ or not is to be determined.

Explanation of Solution

The total pressure of the reaction that is the sum of the partial pressure of water and oxygen is $765\text{ torr}$ . The partial pressure of the water is determined by substituting data from partial pressure table 9.2, that is, $30\text{ torr}$ at $29°\text{C}$ . The partial pressure of oxygen is determined by using Dalton’s law of partial pressure.

$P_{\text{Total}}=P_{\text{Oxygen}}+P_{\text{water}}$

Thus, the partial pressure of oxygen is calculated as shown below:

$\begin{array}{c}765\text{ torr}=P_{\text{Oxygen}}+30\text{ torr}\\ P_{\text{Oxygen}}=765\text{ torr}-30\text{ torr}\\ =735\text{ torr}\end{array}$

So, the partial pressure of oxygen is $735\text{ torr}$ and statement C is correct.

(d)

Interpretation Introduction

Interpretation:

Whether $122.55\text{ g}$ of ${\text{KClO}}_{\text{3}}$ reacts or not is to be determined.

Explanation of Solution

According to the reaction,

$2{\text{ mol KClO}}_{\text{3}}\text{ produces}=3{\text{ mol O}}_{\text{2}}$

$7.2\text{g}$ ${\text{KClO}}_{\text{3}}$ completely reacts in the reaction. So, $122.55\text{ g}$ of ${\text{KClO}}_{\text{3}}$ does not react and statement D is incorrect.

(e)

Interpretation Introduction

Interpretation:

Whether $0.0918\text{ mol}$ of ${\text{KClO}}_{\text{3}}$ reacts or not is to be determined.

Explanation of Solution

According to the reaction,

$\begin{array}{l}2{\text{ mol KClO}}_{\text{3}}\text{ produces}=3{\text{ mol O}}_{\text{2}}\\ 1{\text{ mol KClO}}_{\text{3}}\text{ produces}=\frac{3}{2}{\text{mol O}}_2\end{array}$

The molar mass of ${\text{KClO}}_{\text{3}}$ is $122.6{\text{ g mol}}^{-1}$ . So, $1\text{ mol}$ of ${\text{KClO}}_{\text{3}}$ contains $122.6\text{ g}$ ${\text{KClO}}_{\text{3}}$ .

$\begin{array}{c}1{\text{ mol KClO}}_3\text{ contains}=122.6{\text{ g KClO}}_{\text{3}}\\ 1{\text{ g KClO}}_3=\frac{1}{122.6}{\text{ mol KClO}}_3\\ 7.5{\text{ g KClO}}_3=\frac{1}{122.6}\times 7.5{\text{ mol KClO}}_3\\ =0.0612{\text{ mol KClO}}_{\text{3}}\end{array}$

So, $0.0612{\text{ mol of KClO}}_{\text{3}}$ reacts in the reaction and statement E is incorrect.