Chapter F, Problem M.8E

(a)

Interpretation Introduction

Interpretation:

Balanced equation for the reaction of aluminum oxide with carbon and chlorine to produce aluminum chloride and carbon monoxide has to be written.

Explanation of Solution

The raw chemical equation for the reaction of aluminum oxide with carbon and chlorine to produce aluminum chloride and carbon monoxide is given as follows;

    ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}(s)+\text{C}(s)+{\text{Cl}}_{\text{2}}(g)\to {\text{AlCl}}_{\text{3}}(s)+\text{CO}(g)$

Balancing aluminum atoms:

In the above equation, there are two aluminum atoms on the left side while only one aluminum atom is present in the right side of equation.  Adding coefficient $2$ before ${\text{AlCl}}_{\text{3}}$ in the product side balances the aluminum atom on both sides of the equation.  The chemical equation obtained is given as shown below;

    ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}(s)+\text{C}(s)+{\text{Cl}}_{\text{2}}(g)\to {\text{2AlCl}}_{\text{3}}(s)+\text{CO}(g)$

Balancing oxygen atoms:

In the above equation, there are three oxygen atoms on the left side while only one oxygen atom is present in the right side of equation.  Adding coefficient $3$ before $\text{CO}$ in the product side balances the oxygen atom on both sides of the equation.  The chemical equation obtained is given as shown below;

    ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}(s)+\text{C}(s)+{\text{Cl}}_{\text{2}}(g)\to {\text{2AlCl}}_{\text{3}}(s)+3\text{CO}(g)$

Balancing carbon atoms:

In the above equation, there is one carbon atom on the left side while three carbon atoms are present in the right side of equation.  Adding coefficient $3$ before $\text{C}$ in the reactant side balances the carbon atom on both sides of the equation.  The chemical equation obtained is given as shown below;

    ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}(s)+3\text{C}(s)+{\text{Cl}}_{\text{2}}(g)\to {\text{2AlCl}}_{\text{3}}(s)+3\text{CO}(g)$

Balancing chlorine atoms:

In the above equation, there is two chlorine atoms on the left side while six chlorine atoms are present in the right side of equation.  Adding coefficient $3$ before ${\text{Cl}}_{\text{2}}$ in the reactant side balances the chlorine atom on both sides of the equation.  The balanced chemical equation obtained is given as shown below;

    ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}(s)+3\text{C}(s)+3{\text{Cl}}_{\text{2}}(g)\to {\text{2AlCl}}_{\text{3}}(s)+3\text{CO}(g)$

(b)

Interpretation Introduction

Interpretation:

The mass of aluminum chloride that can be produced by reacting $\text{185}\text{kg}$ of aluminum oxide with $\text{25}\text{kg}$ of carbon and $\text{100}\text{kg}$ of chlorine has to be calculated.

Answer to Problem M.8E

Amount of aluminum chloride produced is $125.3\text{kg}$.

Explanation of Solution

The balanced chemical equation is given as follows;

    ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}(s)+3\text{C}(s)+3{\text{Cl}}_{\text{2}}(g)\to {\text{2AlCl}}_{\text{3}}(s)+3\text{CO}(g)$

Mass of aluminum oxide taken is $\text{185}\text{kg}$, carbon is $\text{25}\text{kg}$, and that of chlorine is $\text{100}\text{kg}$.  The molar mass of aluminum oxide is $101.96\text{g}\cdot {\text{mol}}^{-1}$.  Molar mass of carbon is $12.01\text{g}\cdot {\text{mol}}^{-1}$.  Molar mass of chlorine molecule is $71\text{g}\cdot {\text{mol}}^{-1}$.  The number of moles of aluminum oxide, carbon and chlorine is calculated as follows;

    $\begin{array}{c}{n}_{({\text{Al}}_{\text{2}}{\text{O}}_{\text{3}})}=\frac{185000\text{g}}{101.96\text{g}\cdot {\text{mol}}^{-1}}\\ =1.81\times {10}^3\text{mol}\\ n_{(\text{C})}=\frac{25000\text{g}}{12.01\text{g}\cdot {\text{mol}}^{-1}}\\ =2.08\times {10}^3\text{mol}\\ n_{({\text{Cl}}_{\text{2}})}=\frac{100000\text{g}}{71\text{g}\cdot {\text{mol}}^{-1}}\\ =1.41\times {10}^3\text{mol}\end{array}$

Limiting reactant:

As there are three reactants in the chemical equation, the limiting reactant can be found out by calculating the number of moles of aluminum chloride produced for the individual reactants.

One mole of aluminum oxide produces two moles of aluminum chloride.  Therefore, $1.81\times {10}^3\text{mol}$ will produce;

    $\begin{array}{c}{n}_{({\text{AlCl}}_{\text{3}})}=1.81\times {10}^3\text{mol of}{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\times \frac{\text{2}\text{mol}\text{of}{\text{AlCl}}_{\text{3}}}{\text{1}\text{mol}\text{of}{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}}\\ =3.62\times {10}^3\text{mol of}{\text{AlCl}}_{\text{3}}\end{array}$

Three moles of carbon produces two moles of aluminum chloride.  Therefore, $2.08\times {10}^3\text{mol}$ will produce;

    $\begin{array}{c}{n}_{({\text{AlCl}}_{\text{3}})}=2.08\times {10}^3\text{mol of}\text{C}\times \frac{\text{2}\text{mol}\text{of}{\text{AlCl}}_{\text{3}}}{\text{3}\text{mol}\text{of}\text{C}}\\ =1.39\times {10}^3\text{mol of}{\text{AlCl}}_{\text{3}}\end{array}$

Three moles of molecular chlorine produces two moles of aluminum chloride.  Therefore, $2.08\times {10}^3\text{mol}$ will produce;

    $\begin{array}{c}{n}_{({\text{AlCl}}_{\text{3}})}=1.41\times {10}^3\text{mol of}{\text{Cl}}_{\text{2}}\times \frac{\text{2}\text{mol}\text{of}{\text{AlCl}}_{\text{3}}}{\text{3}\text{mol}\text{of}{\text{Cl}}_{\text{2}}}\\ =0.94\times {10}^3\text{mol of}{\text{AlCl}}_{\text{3}}\end{array}$

The moles of ${\text{AlCl}}_{\text{3}}$ formed is lesser in case of molecular chlorine.  Thus it is the limiting reactant.  Thus the mass of ${\text{AlCl}}_{\text{3}}$ that will be produced can be calculated as shown below;

    $\begin{array}{c}{m}_{({\text{AlCl}}_{\text{3}})}=0.94\times {10}^3\text{mol}\text{of}{\text{Cl}}_{\text{2}}\times 133.34\text{g}\cdot {\text{mol}}^{-1}\\ =125.3\times {10}^3\text{g}\\ =125.3\text{kg}\end{array}$

Thus the amount of aluminum chloride produced is $125.3\text{kg}$.