Chapter 4, Problem 4.103QP

Describe in words how you would do each of the following preparations. Then give the molecular equation for each preparation.

1. a CuCl₂(s) from CuSO₄(s)
2. b Ca(C₂H₃O₂)₂(s) from CaCO₃(s)
3. c NaNO₃(s) from Na₂SO₃(s)
4. d MgCl₂(s) from Mg(OH)₂(s)

Interpretation Introduction

Interpretation:

Preparations of each of the given compound has to be explained and the molecular equations for each reaction has to be found.

Concept introduction:

A chemical equation is the figurative representation of chemical reaction.  In a chemical equation the reactants are in the left side and the products are in the right side.  A balanced chemical equation serves as an easy tool for understanding a chemical reaction.  There are mainly three types of chemical equations, molecular equations, complete ionic equation and net ionic equation.

In molecular equations the reactants and products are represented as molecular substances, even though they exist as ions in solution phase.  The molecular equation for the reaction between $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ and ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ in solution phase is given below.

${\text{Ca(OH)}}_{\text{2(aq)}}{\text{+Na}}_{\text{2}}{\text{CO}}_{\text{3(aq)}}\stackrel{}{\to }{\text{CaCO}}_{\text{3(s)}}{\text{+2NaOH}}_{\text{(aq)}}$

This equation is helpful in understanding the reactants and products involved in the reaction.

In complete ionic equations the electrolytes are represented as its ions.  Soluble compounds exist as ions in solution.  Complete ionic equation is helpful in understanding the reaction at ionic level.  The complete ionic equation for the reaction between $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ and ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is given below.

${\text{Ca}}^{\text{2+}}{}_{\text{(aq)}}{\text{+2OH}}^{\text{-}}{}_{\text{(aq)}}{\text{+2Na}}^{\text{+}}{}_{\text{(aq)}}{\text{+CO}}_{\text{3}}{{}^{\text{2-}}}_{\text{(aq)}}\stackrel{}{\to }{\text{CaCO}}_{\text{3(s)}}{\text{+2Na}}^{\text{+}}{}_{\text{(aq)}}{\text{+2OH}}^{\text{-}}{}_{\text{(aq)}}$

The solid ${\text{CaCO}}_{\text{3}}$ is insoluble and it exist as solid in solution.

In net ionic equations the ions that are common in the reactant and product sides( Spectator ions) are cancelled.  These spectator ions are not participating in the chemical reactions.  The net ionic equation for the reaction between $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ and ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is given below.  As hydroxide ions and sodium ions are common in both the side it is neglected from the equation.

${\text{Ca}}^{\text{2+}}{}_{\text{(aq)}}\text{+}{\text{CO}}_{\text{3}}{{}^{\text{2-}}}_{\text{(aq)}}\stackrel{}{\to }{\text{CaCO}}_{\text{3(s)}}$

Answer to Problem 4.103QP

The molecular equation

${\text{CuSO}}_{\text{4(aq) }}{\text{+ BaCl}}_{\text{2(aq)}}\stackrel{}{\to }{\text{BaSO}}_{\text{4(s) }}{\text{+ CuCl}}_{\text{2(aq)}}$

Explanation of Solution

In order to prepare ${\text{CuCl}}_{\text{2}}$ from ${\text{CuSO}}_4$ we have to take another compound which can exchange its ions with ${\text{CuSO}}_4$.  Let's take ${\text{BaCl}}_{\text{2}}$ to the beaker containing the solution of ${\text{CuSO}}_4$.  The compounds will exchange their ions to form ${\text{BaSO}}_4$ and ${\text{CuCl}}_{\text{2}}$.  ${\text{BaSO}}_4$ is a solid precipitate and it can be filtered from the reaction mixture. The ${\text{CuCl}}_{\text{2}}$ is prepared in its solid from by evaporation of its solution.

The molecular equation for the reaction is given below.

${\text{CuSO}}_{\text{4(aq) }}{\text{+ BaCl}}_{\text{2(aq)}}\stackrel{}{\to }{\text{BaSO}}_{\text{4(s) }}{\text{+ CuCl}}_{\text{2(aq)}}$

Interpretation Introduction

Interpretation:

Preparations of each of the given compound has to be explained and the molecular equations for each reaction has to be found.

Concept introduction:

A chemical equation is the figurative representation of chemical reaction.  In a chemical equation the reactants are in the left side and the products are in the right side.  A balanced chemical equation serves as an easy tool for understanding a chemical reaction.  There are mainly three types of chemical equations, molecular equations, complete ionic equation and net ionic equation.

In molecular equations the reactants and products are represented as molecular substances, even though they exist as ions in solution phase.  The molecular equation for the reaction between $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ and ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ in solution phase is given below.

${\text{Ca(OH)}}_{\text{2(aq)}}{\text{+Na}}_{\text{2}}{\text{CO}}_{\text{3(aq)}}\stackrel{}{\to }{\text{CaCO}}_{\text{3(s)}}{\text{+2NaOH}}_{\text{(aq)}}$

This equation is helpful in understanding the reactants and products involved in the reaction.

In complete ionic equations the electrolytes are represented as its ions.  Soluble compounds exist as ions in solution.  Complete ionic equation is helpful in understanding the reaction at ionic level.  The complete ionic equation for the reaction between $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ and ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is given below.

${\text{Ca}}^{\text{2+}}{}_{\text{(aq)}}{\text{+2OH}}^{\text{-}}{}_{\text{(aq)}}{\text{+2Na}}^{\text{+}}{}_{\text{(aq)}}{\text{+CO}}_{\text{3}}{{}^{\text{2-}}}_{\text{(aq)}}\stackrel{}{\to }{\text{CaCO}}_{\text{3(s)}}{\text{+2Na}}^{\text{+}}{}_{\text{(aq)}}{\text{+2OH}}^{\text{-}}{}_{\text{(aq)}}$

The solid ${\text{CaCO}}_{\text{3}}$ is insoluble and it exist as solid in solution.

In net ionic equations the ions that are common in the reactant and product sides( Spectator ions) are cancelled.  These spectator ions are not participating in the chemical reactions.  The net ionic equation for the reaction between $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ and ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is given below.  As hydroxide ions and sodium ions are common in both the side it is neglected from the equation.

${\text{Ca}}^{\text{2+}}{}_{\text{(aq)}}\text{+}{\text{CO}}_{\text{3}}{{}^{\text{2-}}}_{\text{(aq)}}\stackrel{}{\to }{\text{CaCO}}_{\text{3(s)}}$

Answer to Problem 4.103QP

The molecular equation

${\text{CaCO}}_{\text{3(s)}}{\text{ + 2HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2(aq)}}\stackrel{}{\to }{\text{Ca(C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{\text{)}}_{\text{2(aq)}}{\text{ + CO}}_{\text{2(g)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

Explanation of Solution

In order to prepare calcium acetate $({\text{Ca(C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{\text{)}}_{\text{2}})$ from calcium carbonate ${\text{(CaCO}}_3)$ we can add acetic acid to the beaker containing ${\text{CaCO}}_3$.  Along with the desired product, carbon dioxide and water will also be formed.  ${\text{Ca(C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{\text{)}}_{\text{2}}$ in solid form is obtained by the evaporation of the solution.  Evaporation also removes ${\text{CO}}_{\text{2}}$ and water from the reaction mixture.

The molecular equation for the reaction is given below.

${\text{CaCO}}_{\text{3(s)}}{\text{ + 2HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2(aq)}}\stackrel{}{\to }{\text{Ca(C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{\text{)}}_{\text{2(aq)}}{\text{ + CO}}_{\text{2(g)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

Interpretation Introduction

Interpretation:

Preparations of each of the given compound has to be explained and the molecular equations for each reaction has to be found.

Concept introduction:

A chemical equation is the figurative representation of chemical reaction.  In a chemical equation the reactants are in the left side and the products are in the right side.  A balanced chemical equation serves as an easy tool for understanding a chemical reaction.  There are mainly three types of chemical equations, molecular equations, complete ionic equation and net ionic equation.

In molecular equations the reactants and products are represented as molecular substances, even though they exist as ions in solution phase.  The molecular equation for the reaction between $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ and ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ in solution phase is given below.

${\text{Ca(OH)}}_{\text{2(aq)}}{\text{+Na}}_{\text{2}}{\text{CO}}_{\text{3(aq)}}\stackrel{}{\to }{\text{CaCO}}_{\text{3(s)}}{\text{+2NaOH}}_{\text{(aq)}}$

This equation is helpful in understanding the reactants and products involved in the reaction.

In complete ionic equations the electrolytes are represented as its ions.  Soluble compounds exist as ions in solution.  Complete ionic equation is helpful in understanding the reaction at ionic level.  The complete ionic equation for the reaction between $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ and ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is given below.

${\text{Ca}}^{\text{2+}}{}_{\text{(aq)}}{\text{+2OH}}^{\text{-}}{}_{\text{(aq)}}{\text{+2Na}}^{\text{+}}{}_{\text{(aq)}}{\text{+CO}}_{\text{3}}{{}^{\text{2-}}}_{\text{(aq)}}\stackrel{}{\to }{\text{CaCO}}_{\text{3(s)}}{\text{+2Na}}^{\text{+}}{}_{\text{(aq)}}{\text{+2OH}}^{\text{-}}{}_{\text{(aq)}}$

The solid ${\text{CaCO}}_{\text{3}}$ is insoluble and it exist as solid in solution.

In net ionic equations the ions that are common in the reactant and product sides( Spectator ions) are cancelled.  These spectator ions are not participating in the chemical reactions.  The net ionic equation for the reaction between $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ and ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is given below.  As hydroxide ions and sodium ions are common in both the side it is neglected from the equation.

${\text{Ca}}^{\text{2+}}{}_{\text{(aq)}}\text{+}{\text{CO}}_{\text{3}}{{}^{\text{2-}}}_{\text{(aq)}}\stackrel{}{\to }{\text{CaCO}}_{\text{3(s)}}$

Answer to Problem 4.103QP

The molecular equation

${\text{Na}}_{\text{2}}{\text{SO}}_{\text{3(s)}}{\text{ + 2HNO}}_{\text{3(aq)}}\stackrel{}{\to }{\text{2NaNO}}_{\text{3(aq)}}{\text{ + SO}}_{\text{2(g)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

Explanation of Solution

In order to prepare ${\text{NaNO}}_{\text{3}}$ from ${\text{Na}}_{\text{2}}{\text{SO}}_3$ we can add nitric acid to the beaker containing ${\text{Na}}_{\text{2}}{\text{SO}}_3$.  Along with the desired product, sulphur dioxide and water will also be formed.  ${\text{NaNO}}_{\text{3}}$ in solid form is obtained by the evaporation of the solution.  Evaporation also removes ${\text{SO}}_{\text{2}}$ and water from the reaction mixture.

The molecular equation for the reaction is given below.

${\text{Na}}_{\text{2}}{\text{SO}}_{\text{3(s)}}{\text{ + 2HNO}}_{\text{3(aq)}}\stackrel{}{\to }{\text{2NaNO}}_{\text{3(aq)}}{\text{ + SO}}_{\text{2(g)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

Interpretation Introduction

Interpretation:

Preparations of each of the given compound has to be explained and the molecular equations for each reaction has to be found.

Concept introduction:

A chemical equation is the figurative representation of chemical reaction.  In a chemical equation the reactants are in the left side and the products are in the right side.  A balanced chemical equation serves as an easy tool for understanding a chemical reaction.  There are mainly three types of chemical equations, molecular equations, complete ionic equation and net ionic equation.

In molecular equations the reactants and products are represented as molecular substances, even though they exist as ions in solution phase.  The molecular equation for the reaction between $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ and ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ in solution phase is given below.

${\text{Ca(OH)}}_{\text{2(aq)}}{\text{+Na}}_{\text{2}}{\text{CO}}_{\text{3(aq)}}\stackrel{}{\to }{\text{CaCO}}_{\text{3(s)}}{\text{+2NaOH}}_{\text{(aq)}}$

This equation is helpful in understanding the reactants and products involved in the reaction.

In complete ionic equations the electrolytes are represented as its ions.  Soluble compounds exist as ions in solution.  Complete ionic equation is helpful in understanding the reaction at ionic level.  The complete ionic equation for the reaction between $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ and ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is given below.

${\text{Ca}}^{\text{2+}}{}_{\text{(aq)}}{\text{+2OH}}^{\text{-}}{}_{\text{(aq)}}{\text{+2Na}}^{\text{+}}{}_{\text{(aq)}}{\text{+CO}}_{\text{3}}{{}^{\text{2-}}}_{\text{(aq)}}\stackrel{}{\to }{\text{CaCO}}_{\text{3(s)}}{\text{+2Na}}^{\text{+}}{}_{\text{(aq)}}{\text{+2OH}}^{\text{-}}{}_{\text{(aq)}}$

The solid ${\text{CaCO}}_{\text{3}}$ is insoluble and it exist as solid in solution.

In net ionic equations the ions that are common in the reactant and product sides( Spectator ions) are cancelled.  These spectator ions are not participating in the chemical reactions.  The net ionic equation for the reaction between $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ and ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is given below.  As hydroxide ions and sodium ions are common in both the side it is neglected from the equation.

${\text{Ca}}^{\text{2+}}{}_{\text{(aq)}}\text{+}{\text{CO}}_{\text{3}}{{}^{\text{2-}}}_{\text{(aq)}}\stackrel{}{\to }{\text{CaCO}}_{\text{3(s)}}$

Answer to Problem 4.103QP

The molecular equation

${\text{Mg(OH)}}_{\text{2(s)}}{\text{ + 2HCl}}_{\text{(aq)}}\stackrel{}{\to }{\text{MgCl}}_{\text{2(aq)}}{\text{ + 2H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

Explanation of Solution

In order to prepare ${\text{MgCl}}_{\text{2}}$ from ${\text{Mg(OH)}}_{\text{2}}$, we can add hydrochloric acid to the beaker containing ${\text{Mg(OH)}}_{\text{2}}$.  Along with the desired product water will also be formed.  ${\text{MgCl}}_{\text{2}}$ in solid form is obtained by the evaporation of the solution.

The molecular equation for the reaction is given below.

${\text{Mg(OH)}}_{\text{2(s)}}{\text{ + 2HCl}}_{\text{(aq)}}\stackrel{}{\to }{\text{MgCl}}_{\text{2(aq)}}{\text{ + 2H}}_{\text{2}}{\text{O}}_{\text{(l)}}$