Chapter 21, Problem 21.51QP

Starting with aluminum, describe with balanced equations how you would prepare (a) Al₂Cl₆, (b) Al₂O₃, (c) Al₂(SO₄)₃, (d) NH₄Al(SO₄)₂ · 12H₂O.

Interpretation Introduction

Interpretation:

The balanced equations for the given aluminum compounds has to be described.

Concept introduction:

• There is a law for conversion of mass in a chemical reaction i.e., the mass of total amount of the product should be equal to the total mass of the reactants.
• The concept of writing a balanced chemical equation is depends on conversion of reactants into products.
• First write the reaction from the given information.
• Then count the number of atoms of each element in reactants as well as products.
• Finally obtained values could place it as coefficients of reactants as well as products.

Answer to Problem 21.51QP

The balanced equation for preparation of ${\text{Al}}_{\text{2}}{\text{Cl}}_{\text{6}}$ is,

${\text{2Al}}_{\text{(s)}}{\text{ + 3Cl}}_{\text{2(g)}}\to {\text{Al}}_{\text{2}}{\text{Cl}}_{\text{6(s)}}$

Explanation of Solution

To write the balanced equation for preparation of ${\text{Al}}_{\text{2}}{\text{Cl}}_{\text{6}}$

We can produce ${\text{Al}}_{\text{2}}{\text{Cl}}_{\text{6}}$ by reaction of aluminum with chlorine gas.

${\text{Al}}_{\text{(s)}}{\text{ + Cl}}_{\text{2(g)}}\to {\text{Al}}_{\text{2}}{\text{Cl}}_{\text{6(s)}}$

Here, the reaction is unbalanced. So we need to balance it.  To balance the reaction, calculate the number of atoms present in left side and right side.  Finally, obtained values could place it as coefficients of reactants as well as products.  The balanced equation is

${\text{2Al}}_{\text{(s)}}{\text{ + 3Cl}}_{\text{2(g)}}\to {\text{Al}}_{\text{2}}{\text{Cl}}_{\text{6(s)}}$

Interpretation Introduction

Interpretation:

The balanced equations for the given aluminum compounds has to be described.

Concept introduction:

• There is a law for conversion of mass in a chemical reaction i.e., the mass of total amount of the product should be equal to the total mass of the reactants.
• The concept of writing a balanced chemical equation is depends on conversion of reactants into products.
• First write the reaction from the given information.
• Then count the number of atoms of each element in reactants as well as products.
• Finally obtained values could place it as coefficients of reactants as well as products.

Answer to Problem 21.51QP

The balanced equation for preparation of ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$ is,

${\text{4Al}}_{\text{(s)}}{\text{ + 3O}}_{\text{2(g) }}\to {\text{2Al}}_{\text{2}}{\text{O}}_{\text{3(s)}}$

Explanation of Solution

To write the balanced equation for preparation of ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$

We can produce ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$ by reaction of aluminum with oxygen gas.

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

Here, the reaction is unbalanced. So we need to balance it.  To balance the reaction, calculate the number of atoms present in left side and right side.  Finally, obtained values could place it as coefficients of reactants as well as products.  The balanced equation is

${\text{4Al}}_{\text{(s)}}{\text{ + 3O}}_{\text{2(g) }}\to {\text{2Al}}_{\text{2}}{\text{O}}_{\text{3(s)}}$

Interpretation Introduction

Interpretation:

The balanced equations for the given aluminum compounds has to be described.

Concept introduction:

• There is a law for conversion of mass in a chemical reaction i.e., the mass of total amount of the product should be equal to the total mass of the reactants.
• The concept of writing a balanced chemical equation is depends on conversion of reactants into products.
• First write the reaction from the given information.
• Then count the number of atoms of each element in reactants as well as products.
• Finally obtained values could place it as coefficients of reactants as well as products.

Answer to Problem 21.51QP

The balanced equation for ${\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}$ is,

${\text{2Al}}_{\text{(s)}}{\text{ + 3H}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}\to {\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3(aq)}}{\text{ + 3H}}_{\text{2(g)}}$

Explanation of Solution

To write the balanced equation for ${\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}$

We can produce ${\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}$ by reaction of aluminum with sulfuric acid.

${\text{Al}}_{\text{(s)}}{\text{ + H}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}\to {\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3(aq)}}{\text{ + H}}_{\text{2(g)}}$

Here, the reaction is unbalanced. So we need to balance it.  To balance the reaction, calculate the number of atoms present in left side and right side.  Finally, obtained values could place it as coefficients of reactants as well as products.  The balanced equation is

${\text{2Al}}_{\text{(s)}}{\text{ + 3H}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}\to {\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3(aq)}}{\text{ + 3H}}_{\text{2(g)}}$

Interpretation Introduction

Interpretation:

The balanced equations for the given aluminum compounds has to be described.

Concept introduction:

• There is a law for conversion of mass in a chemical reaction i.e., the mass of total amount of the product should be equal to the total mass of the reactants.
• The concept of writing a balanced chemical equation is depends on conversion of reactants into products.
• First write the reaction from the given information.
• Then count the number of atoms of each element in reactants as well as products.
• Finally obtained values could place it as coefficients of reactants as well as products.

Answer to Problem 21.51QP

The balanced equation for ${\text{NH}}_{\text{4}}{\text{Al(SO}}_{\text{4}}{\text{)}}_{\text{2}}{}_{}$ is ,

${\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3(aq)}}{\text{ + (NH}}_{\text{4}}{\text{)}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}\to {\text{2NH}}_{\text{4}}{\text{Al(SO}}_{\text{4}}{\text{)}}_{\text{2}}$

Explanation of Solution

To write the balanced equation for ${\text{NH}}_{\text{4}}{\text{Al(SO}}_{\text{4}}{\text{)}}_{\text{2}}{}_{}$

We can produce ${\text{NH}}_{\text{4}}{\text{Al(SO}}_{\text{4}}{\text{)}}_{\text{2}}{}_{}$ by reaction of ${\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3(aq)}}$ with ammonium sulfate and after the evaporation process.

${\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3(aq)}}{\text{ + (NH}}_{\text{4}}{\text{)}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}\to {\text{NH}}_{\text{4}}{\text{Al(SO}}_{\text{4}}{\text{)}}_{\text{2}}{}_{\text{(s)}}$

Here, the reaction is unbalanced. So we need to balance it.  To balance the reaction, calculate the number of atoms present in left side and right side.  Finally, obtained values could place it as coefficients of reactants as well as products.  The balanced equation is

${\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3(aq)}}{\text{ + (NH}}_{\text{4}}{\text{)}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}\to {\text{2NH}}_{\text{4}}{\text{Al(SO}}_{\text{4}}{\text{)}}_{\text{2}}$