Chapter 21, Problem 81GQ

(a)

Interpretation Introduction

Interpretation:

To identify the oxidizing agent and reducing agent in the given reaction.

Concept introduction:

An oxidizing agent is a species which reduce other species by gaining electrons in a chemical reaction whereas a reducing agent is a species which loose electrons in the chemical reaction.

The oxidation number of an atom is the charge that atom would have if the compound was composed of ions or the total number of electrons that an atom gain or loss to form a chemical bond with another atom.

Answer to Problem 81GQ

The ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ molecule acts as the oxidizing agent and ${\text{NH}}_{\text{2}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}$ molecule act as the reducing agent.

Explanation of Solution

The balanced chemical equation between ${\text{NH}}_{\text{2}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}$ and ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ is written as,

    ${\text{NH}}_{\text{2}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\left(\text{l}\right)+2{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\left(\text{l}\right)\to 3{\text{N}}_{\text{2}}\left(\text{g}\right)+4{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)+2{\text{CO}}_{\text{2}}\left(\text{g}\right)$

The oxidation number of nitrogen in ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ molecule is $+4$ is reduced to $0$ in ${\text{N}}_{\text{2}}$ molecule. Therefore, ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ is the oxidizing agent.

The oxidation number of nitrogen in ${\text{NH}}_{\text{2}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}$ molecule is $-2$ is oxidised to $0$ in ${\text{N}}_{\text{2}}$ molecule. Therefore, ${\text{NH}}_{\text{2}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}$ is the reducing agent.

(b)

Interpretation Introduction

Interpretation:

To calculate the mass of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ which is required to react with ${\text{NH}}_{\text{2}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}$ and to calculate the mass of each product produced by the reaction of ${\text{NH}}_{\text{2}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}$ with ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$

Concept introduction:

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{=}\frac{\text{Mass}}{\text{Molecular}\text{mass}}\\ \text{Mass}\text{=}\text{Number}\text{of}\text{moles}\text{×}\text{Molecular}\text{mass}\end{array}$

Answer to Problem 81GQ

The mass of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ molecule is reacted with ${\text{NH}}_{\text{2}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}$ is $1.3\times 10^{4}kg$. The mass of water molecule produced from ${\text{NH}}_{\text{2}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}$ is $4.9\times 10^{3}kg$. The mass of ${\text{N}}_{\text{2}}$ molecule produced from ${\text{NH}}_{\text{2}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}$ is $5.7\times 10^{3}kg$. The mass of ${\text{CO}}_{\text{2}}$ molecule produced from ${\text{NH}}_{\text{2}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}$ is $6\times 10^{3}kg$.

Explanation of Solution

The mass of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ which is required to react with ${\text{NH}}_{\text{2}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}$ and the mass of each product produced by the reaction of ${\text{NH}}_{\text{2}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}$ with ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ is calculated below.

Given:

The mass of ${\text{NH}}_{\text{2}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}$ molecule is $4100\text{kg}$.

The balanced chemical equation between ${\text{NH}}_{\text{2}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}$ and ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ is written as,

    ${\text{NH}}_{\text{2}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\left(\text{l}\right)+2{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\left(\text{l}\right)\to 3{\text{N}}_{\text{2}}\left(\text{g}\right)+4{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)+2{\text{CO}}_{\text{2}}\left(\text{g}\right)$ (1)

The number of moles $\left(n_{{\text{NH}}_{\text{2}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}}\right)$ of ${\text{NH}}_{\text{2}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}$ is equal to the mass of ${\text{NH}}_{\text{2}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}$ molecule divided by its molecular weight. The number of moles $\left(n_{{\text{NH}}_{\text{2}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}}\right)$ is calculated as,

$\begin{array}{c}{n}_{{\text{NH}}_{\text{2}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}}=\frac{4100\times {10}^3\text{g}}{60\text{g}\cdot {\text{mol}}^{-1}}\\ =68.33\times {10}^3\text{mol}\end{array}$

From equation (1), one mole of ${\text{NH}}_{\text{2}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}$ reacted with two moles of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ molecules. Therefore, the number of moles $\left(n_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}\right)$ of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ molecule react with $68.33\times {10}^3\text{mol}$ of ${\text{NH}}_{\text{2}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}$ is written as,

$\begin{array}{c}{n}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}=\left(\frac{2\text{mol}}{1\text{mol}}\right)\left(68.33\times {10}^3\text{mol}\right)\\ =136.66\times {10}^3\text{mol}\end{array}$

The mass $\left(m_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}\right)$ of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ molecule is equal to the product of a number of moles of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ molecule and its molecular weight. The mass $\left(m_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}\right)$ of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ molecule is calculated as,

$\begin{array}{c}{m}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}=\left(136.66\times {10}^3\text{mol}\right)\left(92.011\text{g}\cdot {\text{mol}}^{-1}\right)\\ =12.57\times {10}^3\text{g}\\ =1.3\times {10}^4\text{kg}\end{array}$

Therefore, the mass of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ molecule is reacted with ${\text{NH}}_{\text{2}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}$  is $1.3\times {10}^4\text{kg}$.

From equation (1), one mole of ${\text{NH}}_{\text{2}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}$ produced three moles of ${\text{N}}_{\text{2}}$ molecules. Therefore, the number of moles $\left(n_{{\text{N}}_{\text{2}}}\right)$ of ${\text{N}}_{\text{2}}$ molecule produced from $68.33\times {10}^3\text{mol}$ of ${\text{NH}}_{\text{2}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}$  is written as,

$\begin{array}{c}{n}_{{\text{N}}_{\text{2}}}=\left(\frac{3\text{mol}}{1\text{mol}}\right)\left(68.33\times {10}^3\text{mol}\right)\\ =204.99\times {10}^3\text{mol}\end{array}$

The mass $\left(m_{{\text{N}}_{\text{2}}}\right)$ of ${\text{N}}_{\text{2}}$ molecule is equal to the product of a number of moles of ${\text{N}}_{\text{2}}$ molecule and its molecular weight. The mass $\left(m_{{\text{N}}_{\text{2}}}\right)$ of ${\text{N}}_{\text{2}}$ molecule is calculated as,

$\begin{array}{c}{m}_{{\text{N}}_{\text{2}}}=\left(204.99\times {10}^3\text{mol}\right)\left(28\text{g}\cdot {\text{mol}}^{-1}\right)\\ =5.7\times {10}^6\text{g}\\ =5.7\times {10}^3\text{kg}\end{array}$

Therefore, the mass of ${\text{N}}_{\text{2}}$ molecule produced from ${\text{NH}}_{\text{2}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}$ is $5.7\times {10}^3\text{kg}$.

From equation (1) it is clear that, one mole of ${\text{NH}}_{\text{2}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}$ produced four moles of water molecules. Therefore, the number of moles $\left(n_{{\text{H}}_{\text{2}}\text{O}}\right)$ of water molecule reacted with $68.33\times {10}^3\text{mol}$ of ${\text{NH}}_{\text{2}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}$ is written as,

$\begin{array}{c}{n}_{{\text{H}}_{\text{2}}\text{O}}=\left(\frac{4\text{mol}}{1\text{mol}}\right)\left(68.33\times {10}^3\text{mol}\right)\\ =273.32\times {10}^3\text{mol}\end{array}$

The mass $\left(m_{{\text{H}}_{\text{2}}\text{O}}\right)$ of water molecule is equal to the product of a number of moles of the water molecule and its molecular weight. The mass $\left(m_{{\text{H}}_{\text{2}}\text{O}}\right)$ of water molecule is calculated as,

$\begin{array}{c}{m}_{{\text{H}}_{\text{2}}\text{O}}=\left(273.32\times {10}^3\text{mol}\right)\left(18\text{g}\cdot {\text{mol}}^{-1}\right)\\ =4.9\times {10}^6\text{g}\\ =4.9\times {10}^3\text{kg}\end{array}$

Therefore, the mass of water molecule produced from ${\text{NH}}_{\text{2}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}$ is $4.9\times {10}^3\text{kg}$.

From equation (1), one mole of ${\text{NH}}_{\text{2}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}$ produced two moles of ${\text{CO}}_{\text{2}}$ molecules. Therefore, the number of moles $\left(n_{{\text{CO}}_{\text{2}}}\right)$ of ${\text{CO}}_{\text{2}}$ molecule produced from $68.33\times {10}^3\text{mol}$ of ${\text{NH}}_{\text{2}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}$ is written as,

$\begin{array}{c}{n}_{{\text{CO}}_{\text{2}}}=\left(\frac{2\text{mol}}{1\text{mol}}\right)\left(68.33\times {10}^3\text{mol}\right)\\ =136.66\times {10}^3\text{mol}\end{array}$

The mass $\left(m_{{\text{CO}}_{\text{2}}}\right)$ of ${\text{CO}}_{\text{2}}$ molecule is equal to the product of a number of moles of ${\text{CO}}_{\text{2}}$ molecule and its molecular weight. The mass $\left(m_{{\text{CO}}_{\text{2}}}\right)$ of ${\text{CO}}_{\text{2}}$ molecule is calculated as,

$\begin{array}{c}{m}_{{\text{CO}}_{\text{2}}}=\left(136.66\times {10}^3\text{mol}\right)\left(44\text{g}\cdot {\text{mol}}^{-1}\right)\\ =6\times {10}^6\text{g}\\ =6\times {10}^3\text{kg}\end{array}$

Therefore, the mass of ${\text{CO}}_{\text{2}}$ molecule produced from ${\text{NH}}_{\text{2}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}$ is $6\times {10}^3\text{kg}$.