Chapter F, Problem K.22E

(a)

Interpretation Introduction

Interpretation:

The equation given below has to be balanced.

    ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\text{(l)}\to {\text{NH}}_{\text{3}}\text{(g)}+{\text{N}}_{\text{2}}\text{(g)}$

Explanation of Solution

The given chemical equation is written as follows;

    ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\text{(l)}\to {\text{NH}}_{\text{3}}\text{(g)}+{\text{N}}_{\text{2}}\text{(g)}$

Balancing hydrogen atoms:  In the above equation, there are four hydrogen atoms on left side of the equation while three hydrogen atoms are present in right side of the equation.  Adding coefficient $3$ before ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ and $4$ before ${\text{NH}}_{\text{3}}$ balances the hydrogen atoms on both sides of the equation.  This step balances the other atoms also.  The balanced chemical equation is given as follows;

    ${\text{3N}}_{\text{2}}{\text{H}}_{\text{4}}\text{(l)}\to 4{\text{NH}}_{\text{3}}\text{(g)}+{\text{N}}_{\text{2}}\text{(g)}$

(b)

Interpretation Introduction

Interpretation:

Oxidation number of nitrogen in ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$, ${\text{NH}}_{\text{3}}$, and ${\text{N}}_{\text{2}}$ has to be calculated.

Concept Introduction:

Oxidation state of a species is the one that has a specified oxidation number.  Increase in oxidation number corresponds to oxidation while decrease in oxidation number corresponds to reduction.  Rules for assigning oxidation number for an element is given as follows;

• The oxidation number is zero for the element that is present in uncombined state.
• Sum of the oxidation number of the atoms that is present in it is equal to the charge on the species.
• Oxidation number of the element is the charge that is possessed when the more electronegative atom is imagined to be an ion.

Explanation of Solution

Oxidation number of nitrogen in ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$.

The sum of oxidation state of the individual atoms is equal to the total charge.  Therefore, the oxidation number of nitrogen can be found as shown below;

    $\begin{array}{c}2x+[4(+1)]=0\\ 2x+4=0\\ 2x=-4\\ x=-2\end{array}$

Thus the oxidation state of nitrogen in ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ is $-2$.

Oxidation number of nitrogen in ${\text{NH}}_{\text{3}}$.

The sum of oxidation state of the individual atoms is equal to the total charge.  Therefore, the oxidation number of nitrogen can be found as shown below;

    $\begin{array}{c}x+[3(+1)]=0\\ x+3=0\\ x=-3\end{array}$

Thus the oxidation state of nitrogen in ${\text{NH}}_{\text{3}}$ is $-3$.

Oxidation number of nitrogen in ${\text{N}}_{\text{2}}$.

Nitrogen molecule consists only atoms of nitrogen.  Oxidation number of an element that is in its free form is zero.  Therefore, the oxidation state of nitrogen in ${\text{N}}_{\text{2}}$ is zero.

(c)

Interpretation Introduction

Interpretation:

Oxidizing agent and reducing agent has to be identified in the given reaction.

    ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\text{(l)}\to {\text{NH}}_{\text{3}}\text{(g)}+{\text{N}}_{\text{2}}\text{(g)}$

Concept Introduction:

In redox reactions, oxidizing agent is the one that gets reduced by causing oxidation.  These agents can be ions, elements, or even compounds.  In oxidation, the oxidation number increases due to loss of electrons.

In redox reactions, reducing agent is the one that gets oxidized by causing reduction.  These agents can be ions, elements, or even compounds.  In reduction, the oxidation number decreases due to gain of electrons.

Answer to Problem K.22E

Oxidizing agent and reducing agent is ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$.

Explanation of Solution

The given reaction is written as follows;

    ${\text{3N}}_{\text{2}}{\text{H}}_{\text{4}}\text{(l)}\to 4{\text{NH}}_{\text{3}}\text{(g)}+{\text{N}}_{\text{2}}\text{(g)}$

Oxidation number of the atoms present in the above equation is indicated as follows;

From the above equation, it is found that the oxidation state of nitrogen is increased from $-2$ to $0$.  Therefore, oxidation has taken place.  ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ is oxidized.  Therefore, ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ is reducing agent.

The oxidation state of nitrogen decreases from $-2$ to $-3$ as shown in the equation.  Therefore, ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ is reduced.  Therefore, the oxidizing agent is ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$.

(d)

Interpretation Introduction

Interpretation:

Volume of nitrogen gas that will be obtained from $\text{1}\text{.0}\text{L}$ of hydrazine has to be calculated.

    ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\text{(l)}\to {\text{NH}}_{\text{3}}\text{(g)}+{\text{N}}_{\text{2}}\text{(g)}$

Answer to Problem K.22E

Volume of nitrogen is

Explanation of Solution

The balanced chemical equation for the reaction is written as follows;

    ${\text{3N}}_{\text{2}}{\text{H}}_{\text{4}}\text{(l)}\to 4{\text{NH}}_{\text{3}}\text{(g)}+{\text{N}}_{\text{2}}\text{(g)}$

Density of hydrazine is given as $1.004\text{g}\cdot {\text{cm}}^{-3}$.  Volume is given as one liter.  Therefore, the mass of hydrazine is calculated as follows;

    $\begin{array}{c}\text{Mass}=\text{Volume×Density}\\ =1.0\text{L}\times 1.004\text{g}\cdot {\text{cm}}^{-3}\\ =1000{\text{cm}}^{\text{3}}\times 1.004\text{g}\cdot {\text{cm}}^{-3}\\ =1004\text{g}\end{array}$

Molar mass of hydrazine is $32\text{g}\cdot {\text{mol}}^{-1}$.  The number of moles of hydrazine in $\text{1004}\text{g}$ is calculated as follows;

    $\begin{array}{c}n=\frac{m}{M}\\ =\frac{1004\text{g}}{32\text{g}\cdot {\text{mol}}^{-1}}\\ =31.375\text{mol}\end{array}$

Considering the balanced chemical equation, it is found that three moles of hydrazine gives one mole of nitrogen.  Therefore, moles of nitrogen is calculated as shown below;

    $\begin{array}{c}\text{Moles}\text{of}{\text{N}}_{\text{2}}=\text{31}\text{.375}\text{mol}{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\times \frac{\text{1}\text{mol}{\text{N}}_{\text{2}}}{\text{3}\text{mol}{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}}\\ =10.46\text{mol}\end{array}$

It is given that $\text{28}\text{g}$ of nitrogen occupies a volume of $\text{24}\text{L}$.  Therefore, the volume of nitrogen produced is calculated as follows;

    $\begin{array}{c}\text{Volume}\text{of}{\text{N}}_{\text{2}}=\text{10}\text{.46}\text{mol}{\text{N}}_{\text{2}}\times \frac{\text{24}\text{L}}{\text{1}\text{mol}}\\ =251.04\text{L}\\ =251\text{L}\end{array}$

Thus the volume of nitrogen that will be produced is $251\text{L}$.