Chapter 4, Problem 4.82P

(a)

Interpretation Introduction

Interpretation: Oxidizing and reducing agent in below each step is to be determined.

  $\begin{array}{l}Step1.{\text{ 4NH}}_{\text{3}}\left(g\right)+{\text{5O}}_{\text{2}}\left(g\right)\to 4\text{NO}\left(g\right)+{\text{6H}}_{\text{2}}\text{O}\left(l\right)\\ Step2.\text{ 2NO}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)\to 4{\text{NO}}_{\text{2}}\left(g\right)+\text{NO}\left(g\right)\\ Step3.{\text{ 3NO}}_{\text{2}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 2{\text{HNO}}_{\text{3}}\left(l\right)+\text{NO}\left(g\right)\end{array}$

Concept introduction:Change in the oxidation number of a molecule, ion or atom in a reaction is termed as redox reaction. It occurs due to electron transfer between two species.

Oxidizing agent is the substance that gains its electrons. It reduces itself and oxidizes other species in a chemical reaction. A reducing agent is a substance that loses its electrons in the reaction. Thus, reduces others and oxidizes itself.

Explanation of Solution

In step 1:

The expression used for the calculation of oxidation number of nitrogen in ${\text{NH}}_3$ is as follows:

  $\left[3\left(\text{oxidation number of hydrogen}\right)+\left(\text{oxidation number of nitrogen}\right)\right]=\text{0}$

Rearrange the above equation for oxidation number of nitrogen.

  $\text{Oxidation number of nitrogen}=-3\left(\text{oxidation number of hydrogen}\right)$

The oxidation number of hydrogen is $+1$ .

Substitute the value in above equation.

  $\begin{array}{c}\text{Oxidation number of nitrogen}=-3\left(\text{oxidation number of hydrogen}\right)\\ =-3\left(\text{+1}\right)\\ =-3\end{array}$

The expression used for the calculation of oxidation number of nitrogen in $\text{NO}$ is as follows:

  $\left[\left(\text{oxidation number of oxygen}\right)+\left(\text{oxidation number of nitrogen}\right)\right]=\text{0}$

Rearrange the above equation for oxidation number of nitrogen.

  $\text{Oxidation number of nitrogen}=-\left(\text{oxidation number of oxygen}\right)$

The oxidation number of oxygen is $-2$ .

Substitute the value in above equation.

  $\begin{array}{c}\text{Oxidation number of nitrogen}=-\left(\text{oxidation number of oxygen}\right)\\ =-\left(-\text{2}\right)\\ =+2\end{array}$

Since in step 1 oxidation number of nitrogen increases from $-3$ to $+2$ . Hence, ${\text{NH}}_3$ is oxidized, therefore, act as a reducing agent. And ${\text{O}}_2$ is reduced and act as oxidizing agent.

In step 2:

The expression used for the calculation of oxidation number of nitrogen in $\text{NO}$ is as follows:

  $\left[\left(\text{oxidation number of oxygen}\right)+\left(\text{oxidation number of nitrogen}\right)\right]=\text{0}$

Rearrange the above equation for oxidation number of nitrogen.

  $\text{Oxidation number of nitrogen}=-\left(\text{oxidation number of oxygen}\right)$

The oxidation number of oxygen is $-2$ .

Substitute the value in above equation.

  $\begin{array}{c}\text{Oxidation number of nitrogen}=-\left(\text{oxidation number of oxygen}\right)\\ =-\left(-\text{2}\right)\\ =+2\end{array}$

The expression used for the calculation of oxidation number of nitrogen in ${\text{NO}}_2$ is as follows:

  $\left[2\left(\text{oxidation number of oxygen}\right)+\left(\text{oxidation number of nitrogen}\right)\right]=\text{0}$

Rearrange the above equation for oxidation number of nitrogen.

  $\text{Oxidation number of nitrogen}=-2\left(\text{oxidation number of oxygen}\right)$

The oxidation number of oxygen is $-2$ .

Substitute the value in above equation.

  $\begin{array}{c}\text{Oxidation number of nitrogen}=-2\left(\text{oxidation number of oxygen}\right)\\ =-2\left(-\text{2}\right)\\ =+4\end{array}$

Since in step 2 oxidation number of nitrogen increases from $+2$ to $+4$ . Hence, $\text{NO}$ is oxidized and act as a reducing agent. Whereas ${\text{O}}_2$ is reduced and act as oxidizing agent.

In step 3:

The expression used for the calculation of oxidation number of nitrogen in ${\text{NO}}_2$ is as follows:

  $\left[2\left(\text{oxidation number of oxygen}\right)+\left(\text{oxidation number of nitrogen}\right)\right]=\text{0}$

Rearrange the above equation for oxidation number of nitrogen.

  $\text{Oxidation number of nitrogen}=-2\left(\text{oxidation number of oxygen}\right)$

The oxidation number of oxygen is $-2$ .

Substitute the value in above equation.

  $\begin{array}{c}\text{Oxidation number of nitrogen}=-2\left(\text{oxidation number of oxygen}\right)\\ =-2\left(-\text{2}\right)\\ =+4\end{array}$

The expression used for the calculation of oxidation number of nitrogen in ${\text{HNO}}_3$ is as follows:

  $\left[\begin{array}{l}\left(\text{oxidation number of hydrogen}\right)+\left(\text{oxidation number of nitrogen}\right)+\\ 3\left(\text{oxidation number of oxygen}\right)\end{array}\right]=\text{0}$

Rearrange the above equation for oxidation number of nitrogen.

  $\text{Oxidation number of nitrogen}=\left[\begin{array}{l}-3\left(\text{oxidation number of oxygen}\right)\\ -1\left(\text{oxidation number of hydrogen}\right)\end{array}\right]$

The oxidation number of oxygen is $-2$ .

The oxidation number of hydrogen is $+1$ .

Substitute the value in above equation.

  $\begin{array}{c}\text{Oxidation number of nitrogen}=\left[\begin{array}{l}-3\left(\text{oxidation number of oxygen}\right)\\ -1\left(\text{oxidation number of hydrogen}\right)\end{array}\right]\\ =\left[-3\left(-\text{2}\right)-1\left(+1\right)\right]\\ =\left[6-1\right]\\ =+5\end{array}$

Since in step 3 oxidation number of nitrogen increases from $+4$ in ${\text{NO}}_2$ to $+5$ in ${\text{HNO}}_3$ . Hence, ${\text{NO}}_2$ is oxidized and acts as a reducing agent. Whereas ${\text{NO}}_2$ is reduced to $\text{NO}$ and acts as oxidizing agent.

Therefore, in step 1, ${\text{O}}_2$ acts as oxidizing agent and ${\text{NH}}_3$ as reducing agent.In step 2, ${\text{O}}_2$ acts as oxidizing agent and $\text{NO}$ as reducing agent. In step 3, ${\text{NO}}_2$ acts as oxidizing agent and ${\text{NO}}_2$ as reducing agent.

(b)

Interpretation Introduction

Interpretation: Mass of ammonia used to produce $3.0\times {10}^4\text{kg}$ of ${\text{HNO}}_3$ is to be calculated.

Concept introduction:Change in the oxidation number of a molecule, ion or atom in a reaction is termed as redox reaction. It occurs due to electron transfer between two species.

Oxidizing agent is the substance that gains its electrons. It reduces itself and oxidizes other species in a chemical reaction. A reducing agent is a substance that loses its electrons in the reaction. Thus, reduces others and oxidizes itself.

Answer to Problem 4.82P

Mass of ammonia used to produce $3.0\times {10}^4\text{kg}$ of ${\text{HNO}}_3$ is $1.2\times {10}^4\text{kg}$ .

Explanation of Solution

Three moles of ${\text{NO}}_2$ reacts to form two moles of ${\text{HNO}}_3$ . The reaction of ${\text{NO}}_2$ to give ${\text{HNO}}_3$ is as follows:

  ${\text{3NO}}_{\text{2}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 2{\text{HNO}}_{\text{3}}\left(l\right)+\text{NO}\left(g\right)$

The formula used for the calculation of moles of ${\text{NO}}_2$ is as follows:

  $\text{Moles}\text{of}{\text{NO}}_2=\left[\left(\frac{\text{mass of}{\text{HNO}}_3\left(\text{g}\right)}{\text{molecular mass}\text{of}{\text{HNO}}_3\left(\text{g/mol}\right)}\right)\left(\frac{3\text{mol}{\text{NO}}_2}{2\text{mol}{\text{HNO}}_3}\right)\right]$

The mass of ${\text{HNO}}_3$ is $3.0\times {10}^4\text{kg}$ .

Molecular mass of ${\text{HNO}}_3$ is $63.02\text{g/mol}$ .

Substitute the values in above equation.

  $\begin{array}{c}\text{Moles}\text{of}{\text{NO}}_2=\left[\left(\frac{\text{mass of}{\text{HNO}}_3\left(\text{g}\right)}{\text{molecular mass}\text{of}{\text{HNO}}_3\left(\text{g/mol}\right)}\right)\left(\frac{3\text{mol}{\text{NO}}_2}{2\text{mol}{\text{HNO}}_3}\right)\right]\\ =\left[\left(\frac{3.0\times {10}^4\text{kg}}{63.02\text{g/mol}}\right)\left(\frac{1000\text{g}}{1\text{kg}}\right)\left(\frac{3\text{mol}{\text{NO}}_2}{2\text{mol}{\text{HNO}}_3}\right)\right]\\ =\left[\left(47.60\times {10}^4\text{mol}\right)\left(\frac{3\text{mol}{\text{NO}}_2}{2\text{mol}{\text{HNO}}_3}\right)\right]\\ =7.14059\times {10}^5\text{mol}\end{array}$

Two moles of $\text{NO}$ reacts to form four moles of ${\text{NO}}_2$ . The reaction of $\text{NO}$ to give ${\text{NO}}_2$ is as follows:

  $\text{2NO}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)\to 4{\text{NO}}_{\text{2}}\left(g\right)+\text{NO}\left(g\right)$

The formula used for the calculation of moles of $\text{NO}$ is as follows:

  $\text{Moles}\text{of}\text{NO}=\left[\left({\text{moles of NO}}_2\right)\left(\frac{2\text{mol}\text{NO}}{2\text{mol}{\text{NO}}_2}\right)\right]$

Moles of ${\text{NO}}_2$ is $7.14059\times {10}^5\text{mol}$ .

Substitute the values in above equation.

  $\begin{array}{c}\text{Moles}\text{of}\text{NO}=\left[\left({\text{moles of NO}}_2\right)\left(\frac{2\text{mol}\text{NO}}{2\text{mol}{\text{NO}}_2}\right)\right]\\ =\left[\left(7.14059\times {10}^5\text{mol}\right)\left(\frac{2\text{mol}\text{NO}}{2\text{mol}{\text{NO}}_2}\right)\right]\\ =7.14059\times {10}^5\text{mol}\end{array}$

Four moles of ${\text{NH}}_3$ reacts to form four moles of $\text{NO}$ . The reaction of ${\text{NH}}_3$ to give $\text{NO}$ is as follows:

  ${\text{4NH}}_{\text{3}}\left(g\right)+{\text{5O}}_{\text{2}}\left(g\right)\to 4\text{NO}\left(g\right)+{\text{6H}}_{\text{2}}\text{O}\left(l\right)$

The formula used for the calculation of moles of ${\text{NH}}_3$ is as follows:

  $\text{Moles}\text{of}{\text{NH}}_3=\left[\left(\text{moles of NO}\right)\left(\frac{4\text{mol}{\text{NH}}_3}{4\text{mol}\text{NO}}\right)\right]$

Moles of $\text{NO}$ is $7.14059\times {10}^5\text{mol}$ .

Substitute the values in above equation.

  $\begin{array}{c}\text{Moles}\text{of}{\text{NH}}_3=\left[\left(\text{moles of NO}\right)\left(\frac{4\text{mol}{\text{NH}}_3}{4\text{mol}\text{NO}}\right)\right]\\ =\left[\left(7.14059\times {10}^5\text{mol}\right)\left(\frac{4\text{mol}{\text{NH}}_3}{4\text{mol}\text{NO}}\right)\right]\\ =7.14059\times {10}^5\text{mol}\end{array}$

The formula used for the calculation of mass of ${\text{NH}}_3$ is as follows:

  $\text{Mass}\text{of}{\text{NH}}_3=\left(\text{moles of}{\text{NH}}_3\right)\left({\text{molecular mass of NH}}_3\right)$

Moles of ${\text{NH}}_3$ is $7.14059\times {10}^5\text{mol}$ .

Molecular mass of ${\text{NH}}_3$ is $17.03\text{g/mol}$ .

Substitute the values in above equation.

  $\begin{array}{c}\text{Mass}\text{of}{\text{NH}}_3=\left(\text{moles of}{\text{NH}}_3\right)\left({\text{molecular mass of NH}}_3\right)\\ =\left(7.14059\times {10}^5\text{mol}\right)\left(17.03\text{g/mol}\right)\left(\frac{1\text{kg}}{1000\text{g}}\right)\\ =1.21604\times {10}^4\text{kg}\\ \approx 1.2\times {10}^4\text{kg}\end{array}$

Hence, mass of ammonia, ${\text{NH}}_3$ is $1.2\times {10}^4\text{kg}$ .