Chapter 4, Problem 4.155P

(a)

Interpretation Introduction

Interpretation:

The balanced equation for the reaction is to be written. Also, the reducing and oxidizing agent is to be identified.

Concept introduction:

A redox reaction is a type of reaction that involves the change in oxidation number of a molecule, atom or ion changes due to the transfer of an electron from one species to another.

An oxidizing agent is a substance that oxidizes another species and itself gets reduced in a chemical reaction. A reducing agent is the one that reduces another species and itself gets oxidized in a chemical reaction.

Answer to Problem 4.155P

The balanced equation for the reaction is as follows:

  ${\text{3NH}}_{\text{4}}{\text{ClO}}_{\text{4}}\left(s\right)+3\text{Al}\left(aq\right)\stackrel{\text{Sunlight}}{\to }{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+{\text{AlCl}}_3\left(s\right)+6{\text{H}}_{\text{2}}\text{O}\left(g\right)+3\text{NO}\left(g\right)$

The reducing agent is ${\text{NH}}_{\text{4}}{\text{ClO}}_{\text{4}}$ and the oxidizing agent is $\text{Al}$.

Explanation of Solution

The given balanced chemical reaction is as follows:

  ${\text{3NH}}_{\text{4}}{\text{ClO}}_{\text{4}}\left(s\right)+3\text{Al}\left(aq\right)\stackrel{\text{Sunlight}}{\to }{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+{\text{AlCl}}_3\left(s\right)+6{\text{H}}_{\text{2}}\text{O}\left(g\right)+3\text{NO}\left(g\right)$

The oxidation number of aluminium is $+3$, the oxidation number of hydrogen attached to non-metal is $+1$ and oxidation number of nitrogen in ammonium ion is $-3$.

The expression to calculate the oxidation number of chlorine in ${\text{NH}}_{\text{4}}{\text{ClO}}_{\text{4}}$ is,

  $\left[\begin{array}{l}\left(\text{oxidation number of nitrogen}\right)\\ +4\left(\text{oxidation number of hydrogen}\right)\\ +4\left(\text{oxidation number of oxygen}\right)\\ +\left(\text{oxidation number of chloride}\right)\end{array}\right]=0$        (1)

Rearrange equation (1) for the oxidation number of chlorine.

  $\text{Oxidation number of chloride}=\left[\begin{array}{l}-\left(\text{oxidation number of nitrogen}\right)\\ -4\left(\text{oxidation number of hydrogen}\right)\\ -4\left(\text{oxidation number of oxygen}\right)\end{array}\right]$        (2)

Substitute $-2$ for the oxidation number of oxygen, $-3$ for the oxidation number of nitrogen and $+1$ for the oxidation number of hydrogen in equation (2).

  $\begin{array}{c}\text{Oxidation number of chloride}=\left[-\left(-3\right)-4\left(+1\right)-4\left(-2\right)\right]\\ =\left[+3-4+8\right]\\ =+7\end{array}$

The expression to calculate the oxidation number of chlorine in ${\text{AlCl}}_3$ is as follows:

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

Rearrange equation (3) for the oxidation number of chlorine.

  $\text{Oxidation number of chlorine}=\frac{1}{3}\left[-\left(\text{oxidation number of aluminium}\right)\right]$        (4)

Substitute $+3$ for the oxidation number of aluminium in equation (4).

  $\begin{array}{c}\text{Oxidation number of chlorine}=\frac{1}{3}\left[-\left(+3\right)\right]\\ =-1\end{array}$

The oxidation state of chlorine in ${\text{NH}}_{\text{4}}{\text{ClO}}_{\text{4}}$ is $+7$ and it decreases to $-1$ in ${\text{AlCl}}_3$. ${\text{NH}}_{\text{4}}{\text{ClO}}_{\text{4}}$ undergoes reduction in the reaction, therefore, it acts as the oxidizing agent.

The oxidation state of aluminium in $\text{Al}$ is 0 and it increases to $+3$ in ${\text{AlCl}}_3$. $\text{Al}$ undergoes oxidation in the reaction, therefore, it acts as the reducing agent.

Conclusion

An oxidizing agent is a substance that oxidizes another species and itself gets reduced in a chemical reaction. A reducing agent is the one that reduces another species and itself gets oxidized in a chemical reaction.

(b)

Interpretation Introduction

Interpretation:

The total mole of gas produced when $50.0\text{ kg}$ of ammonium perchlorate reacts with a stoichiometric amount of Al is to be calculated.

Concept introduction:

The expression to calculate the moles of solute when given mass and molecular mass of compound are given is as follows:

  $\text{Moles of compound}\left(\text{mol}\right)=\left[\begin{array}{c}\text{given mass}\text{of compound}\left(\text{g}\right)\\ \left(\frac{\text{1mole}\text{of compound}\left(\text{mol}\right)}{\text{molecular mass of compound}\left(\text{g}\right)}\right)\end{array}\right]$

Stoichiometry of a reaction is utilized to determine the amount of any species in the reaction by the relationship between the reactants and products.

Consider the general reaction,

$\text{A}+2\text{B}\to 3\text{C}$

One mole of $\text{A}$ reacts with two moles of $\text{B}$ to produce three moles of $\text{C}$. The stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3 and the stoichiometric ratio between B and C is 2:3.

Answer to Problem 4.155P

$1.28\times {10}^3\text{mol}$ of gas is produced when $50.0\text{ kg}$ of ammonium perchlorate reacts with a stoichiometric amount of Al.

Explanation of Solution

The given balanced chemical reaction is as follows:

  ${\text{3NH}}_{\text{4}}{\text{ClO}}_{\text{4}}\left(s\right)+3\text{Al}\left(aq\right)\stackrel{\text{Sunlight}}{\to }{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+{\text{AlCl}}_3\left(s\right)+6{\text{H}}_{\text{2}}\text{O}\left(g\right)+3\text{NO}\left(g\right)$

The formula to calculate the moles of ${\text{NH}}_{\text{4}}{\text{ClO}}_{\text{4}}$ is as follows:

  ${\text{Moles of NH}}_{\text{4}}{\text{ClO}}_{\text{4}}=\left[\frac{\text{given mass}{\text{of NH}}_{\text{4}}{\text{ClO}}_{\text{4}}}{{\text{molecular mass of NH}}_{\text{4}}{\text{ClO}}_{\text{4}}}\right]$        (5)

Substitute $50.0\text{ kg}$ for the mass of ${\text{NH}}_{\text{4}}{\text{ClO}}_{\text{4}}$ and $117.49\text{ g/mol}$ for the molar mass of ${\text{NH}}_{\text{4}}{\text{ClO}}_{\text{4}}$ in the equation (5).

  $\begin{array}{c}{\text{Moles of NH}}_{\text{4}}{\text{ClO}}_{\text{4}}=\left(\frac{50.0\text{ kg}}{117.49\text{ g/mol}}\right)\left(\frac{1000\text{g}}{\text{1}\text{kg}}\right)\\ =425.568133\text{mol}\end{array}$

The formula to calculate the moles of gas is as follows:

  $\text{Moles of}\text{gas}=\text{moles of}{\text{NH}}_{\text{4}}{\text{ClO}}_{\text{4}}\left(\frac{9\text{mol}\text{gas}}{3\text{mol}{\text{NH}}_{\text{4}}{\text{ClO}}_{\text{4}}}\right)$        (6)

Substitute $425.568133\text{mol}$ for moles of ${\text{NH}}_{\text{4}}{\text{ClO}}_{\text{4}}$ in the equation (6).

  $\begin{array}{c}\text{Moles of}\text{gas}=\left(425.568133\text{mol}\right)\left(\frac{9\text{mol}\text{gas}}{3\text{mol}{\text{NH}}_{\text{4}}{\text{ClO}}_{\text{4}}}\right)\\ =1276.70\text{mol}\\ \approx 1.28\times {10}^3\text{mol}\text{.}\end{array}$

Conclusion

Stoichiometry of a reaction can be used to determine the moles of a chemical species when the moles of other chemical species present in the reaction is given.

(c)

Interpretation Introduction

Interpretation:

The change in volume from the reaction is to be calculated.

Concept introduction:

Density is defined as mass per unit volume. Mass and volume are physical quantities and the units of mass and volume are fundamental units. Density is the ratio of mass to the volume. The unit of volume is derived from the units of mass and volume. The SI unit of density is ${\text{kg/m}}^{\text{3}}$. The formula to calculate density is,

  $\text{Density}=\frac{\text{Mass}}{\text{Volume}}$        (7)

Stoichiometry of a reaction is utilized to determine the amount of any species in the reaction by the relationship between the reactants and products.

Consider the general reaction,

  $\text{A}+2\text{B}\to 3\text{C}$

One mole of $\text{A}$ reacts with two moles of $\text{B}$ to produce three moles of $\text{C}$. The stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3 and the stoichiometric ratio between B and C is 2:3.

The conversion factor to convert $\text{mL}$ to $\text{cc}$ is as follows:

  $1\text{mL}=\text{1}\text{cc}$

Answer to Problem 4.155P

The change in volume is $2.86\times {10}^4\text{L}$.

Explanation of Solution

The given balanced chemical reaction is as follows:

  ${\text{3NH}}_{\text{4}}{\text{ClO}}_{\text{4}}\left(s\right)+3\text{Al}\left(aq\right)\stackrel{\text{Sunlight}}{\to }{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+{\text{AlCl}}_3\left(s\right)+6{\text{H}}_{\text{2}}\text{O}\left(g\right)+3\text{NO}\left(g\right)$

Rearrange the equation (7) to calculate the volume.

  $\text{Volume}=\frac{\text{Mass}}{\text{Density}}$        (8)

Substitute $1.9\text{g/cc}$ for density and $50.0\text{ kg}$ for mass in the equation (8) to calculate the volume of ${\text{NH}}_{\text{4}}{\text{ClO}}_{\text{4}}$.

  $\begin{array}{c}\text{Volume}{\text{of NH}}_{\text{4}}{\text{ClO}}_{\text{4}}=\left(\frac{50.0\text{ kg}}{1.9\text{g/cc}}\right)\left(\frac{1000\text{g}}{\text{1}\text{kg}}\right)\left(\frac{1\text{mL}}{\text{1}\text{cc}}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\\ =25.6410\text{L}\end{array}$

The formula to calculate the mass of $\text{Al}$ is as follows:

  $\text{Mass of}\text{Al}=\text{moles of}{\text{NH}}_{\text{4}}{\text{ClO}}_{\text{4}}\left(\frac{3\text{mol}\text{Al}}{3\text{mol}{\text{NH}}_{\text{4}}{\text{ClO}}_{\text{4}}}\right)\left(\text{molar mass of Al}\right)$        (9)

Substitute $425.568133\text{mol}$ for moles of ${\text{NH}}_{\text{4}}{\text{ClO}}_{\text{4}}$ and $26.98\text{ g/mol}$ for molar mass of $\text{Al}$ in the equation (9).

  $\begin{array}{c}\text{Mass of}\text{Al}=425.568133\text{mol}\left(\frac{3\text{mol}\text{Al}}{3\text{mol}{\text{NH}}_{\text{4}}{\text{ClO}}_{\text{4}}}\right)\left(26.98\text{ g/mol}\right)\\ =11481.828\text{g}\end{array}$

Substitute $2.70\text{g/cc}$ for density and $11481.828\text{g}$ for mass in the equation (8) to calculate the volume of $\text{Al}$.

  $\begin{array}{c}\text{Volume}\text{of Al}=\left(\frac{11481.828\text{g}}{2.70\text{g/cc}}\right)\left(\frac{1\text{mL}}{\text{1}\text{cc}}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\\ =4.2525\text{L}\end{array}$

The formula to calculate the initial volume is as follows:

  $\text{Initial Volume}={\text{volume of NH}}_{\text{4}}{\text{ClO}}_{\text{4}}+\mathrm{volume}of\text{Al}$        (10)

Substitute $25.6410\text{L}$ for volume of ${\text{NH}}_{\text{4}}{\text{ClO}}_{\text{4}}$ and $4.2525\text{L}$ for volume of $\text{Al}$ in the equation (10).

  $\begin{array}{c}\text{Initial Volume}=25.6410\text{L}+4.2525\text{L}\\ =29.8935\text{ L}\end{array}$

The formula to calculate the mass of ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$ is as follows:

  $\text{Mass of}{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}=\left[\begin{array}{l}\text{moles of}{\text{NH}}_{\text{4}}{\text{ClO}}_{\text{4}}\left(\frac{1\text{mol}{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}}{3\text{mol}{\text{NH}}_{\text{4}}{\text{ClO}}_{\text{4}}}\right)\\ \left({\text{molar mass of Al}}_{\text{2}}{\text{O}}_{\text{3}}\right)\end{array}\right]$        (11)

Substitute $425.568133\text{mol}$ for moles of ${\text{NH}}_{\text{4}}{\text{ClO}}_{\text{4}}$ and $101.96\text{ g/mol}$ for molar mass of ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$ in the equation (11).

  $\begin{array}{c}\text{Mass of}{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}=\left[\begin{array}{l}\left(425.568133\text{mol}\right)\left(\frac{1\text{mol}{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}}{3\text{mol}{\text{NH}}_{\text{4}}{\text{ClO}}_{\text{4}}}\right)\\ \left(101.96\text{ g/mol}\right)\end{array}\right]\\ =14463.64\text{g}\end{array}$

Substitute $3.97\text{g/cc}$ for density and $14463.64\text{g}$ for mass in the equation (8) to calculate the volume of ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$.

$\begin{array}{c}\text{Volume}{\text{of Al}}_{\text{2}}{\text{O}}_{\text{3}}=\left(\frac{14463.64\text{g}}{3.97\text{g/cc}}\right)\left(\frac{1\text{mL}}{\text{1}\text{cc}}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\\ =3.6432\text{L}\end{array}$

The formula to calculate the mass of ${\text{AlCl}}_3$ is as follows:

  $\text{Mass of}{\text{AlCl}}_3=\left[\begin{array}{l}\text{moles of}{\text{NH}}_{\text{4}}{\text{ClO}}_{\text{4}}\left(\frac{1\text{mol}{\text{AlCl}}_3}{3\text{mol}{\text{NH}}_{\text{4}}{\text{ClO}}_{\text{4}}}\right)\\ \left({\text{molar mass of AlCl}}_3\right)\end{array}\right]$        (12)

Substitute $425.568133\text{mol}$ for moles of ${\text{NH}}_{\text{4}}{\text{ClO}}_{\text{4}}$ and $133.33\text{ g/mol}$ for molar mass of ${\text{AlCl}}_3$ in the equation (12).

  $\begin{array}{c}\text{Mass of}{\text{AlCl}}_3=\left[\begin{array}{l}\left(425.568133\text{mol}\right)\left(\frac{1\text{mol}{\text{AlCl}}_3}{3\text{mol}{\text{NH}}_{\text{4}}{\text{ClO}}_{\text{4}}}\right)\\ \left(133.33\text{ g/mol}\right)\end{array}\right]\\ =18913.67\text{g}\end{array}$

Substitute $2.44\text{g/cc}$ for density and $18913.67\text{g}$ for mass in the equation (8) to calculate the volume of ${\text{AlCl}}_3$.

  $\begin{array}{c}\text{Volume}{\text{of AlCl}}_3=\left(\frac{18913.67\text{g}}{2.44\text{g/cc}}\right)\left(\frac{1\text{mL}}{\text{1}\text{cc}}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\\ =7.7515\text{L}\end{array}$

The volume of one mole of gas at STP is $22.4\text{L}$. The volume of $1276.70\text{mol}$ of gas is calculated as follows:

  $\begin{array}{c}\text{Volume of}\text{gas}=\left(1276.70\text{mol}\right)\left(\frac{22.4\text{L}}{\text{1}\text{mol}}\right)\\ =28598.08\text{L}\end{array}$

The formula to calculate the final volume is as follows:

  $\text{Final Volume}={\text{volume of Al}}_{\text{2}}{\text{O}}_{\text{3}}+\mathrm{volume}of{\text{AlCl}}_3+\mathrm{volume}of\text{gas}$        (13)

Substitute $3.6432\text{L}$ for volume of ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$, $7.7515\text{L}$ for volume of ${\text{AlCl}}_3$ and $28598.08\text{L}$ for volume of gas in the equation (13).

  $\begin{array}{c}\text{Final Volume}=3.6432\text{L}+7.7515\text{L}+28598.08\text{L}\\ =28609.4747\text{L}\end{array}$

The formula to calculate the volume change is as follows:

  $\text{Volume change}=\text{Final Volume}-\text{Initial Volume}$        (14)

Substitute $28609.4747\text{L}$ for final volume and $29.8935\text{ L}$ for initial volume in the equation (14).

  $\begin{array}{c}\text{Volume change}=28609.4747\text{L}-29.8935\text{ L}\\ =28579.5812\text{L}\\ \approx 2.86\times {10}^4\text{L}\text{.}\end{array}$

Conclusion

The volume change in a reaction is defined as the difference between volume of reactants and the volume of products.