Chapter F, Problem I.15E

(a)

Interpretation Introduction

Interpretation:

Chemical equation, overall equation, complete ionic equation and also the net ionic equation for the reaction between ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{CrO}}_{\text{4}}$ and ${\text{BaCl}}_{\text{2}}$ has to be written.

Explanation of Solution

The reaction given is between aqueous solutions of ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{CrO}}_{\text{4}}$ and ${\text{BaCl}}_{\text{2}}$.  The reaction can be given as follows:

    ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{CrO}}_{\text{4}}(\text{aq})+{\text{BaCl}}_{\text{2}}(\text{aq})\to {\text{BaCrO}}_{\text{4}}(s)+{\text{2NH}}_{\text{4}}\text{Cl(aq)}$

Aqueous solution of ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{CrO}}_{\text{4}}$ reacts with ${\text{BaCl}}_{\text{2}}$ resulting in formation of ${\text{BaCrO}}_{\text{4}}$ and ${\text{NH}}_{\text{4}}\text{Cl}$.  The precipitate that is formed is ${\text{BaCrO}}_{\text{4}}$.  This is because the chromate salts of ${\text{Ca}}^{\text{2+}}$, ${\text{Sr}}^{\text{2+}}$, ${\text{Ba}}^{\text{2+}}$, ${\text{Pb}}^{\text{2+}}$, ${\text{Hg}}_{\text{2}}{}^{\text{2+}}$, and ${\text{Ag}}^{\text{+}}$ are alone insoluble.

The balanced chemical equation for the reaction is given as follows:

    ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{CrO}}_{\text{4}}(\text{aq})+{\text{BaCl}}_{\text{2}}(\text{aq})\to {\text{BaCrO}}_{\text{4}}(s)+{\text{2NH}}_{\text{4}}\text{Cl(aq)}$

Complete ionic equation is written by showing the ions present in aqueous solution.  This is given as follows:

${\text{2NH}}_{\text{4}}{}^{\text{+}}{\text{(aq)+CrO}}_{\text{4}}{}^{2-}(\text{aq})+{\text{Ba}}^{\text{2+}}{\text{(aq)+2Cl}}^{-}(\text{aq})\to {\text{BaCrO}}_{\text{4}}(s)+2{\text{NH}}_{\text{4}}{}^{\text{+}}{\text{(aq)+2Cl}}^{-}\text{(aq)}$

In the complete ionic equation, the common ions that are present on both sides of the equation is known as spectator ions.  Thus in this case, the spectator ions are ${\text{NH}}_{\text{4}}{}^{\text{+}}$ and ${\text{Cl}}^{-}$.  Net ionic equation can be obtained from the complete ionic equation by removing the spectator ions.

$\cancel{{\text{2NH}}_{\text{4}}{}^{\text{+}}\text{(aq)}}{\text{+CrO}}_{\text{4}}{}^{2-}(\text{aq})+{\text{Ba}}^{\text{2+}}\text{(aq)+}\cancel{{\text{2Cl}}^{-}(\text{aq})}\to {\text{BaCrO}}_{\text{4}}(s)+\cancel{2{\text{NH}}_{\text{4}}{}^{\text{+}}\text{(aq)}}\text{+}\cancel{{\text{2Cl}}^{-}\text{(aq)}}$

Thus net ionic equation can be written as follows:

    ${\text{CrO}}_{\text{4}}{}^{2-}(\text{aq})+{\text{Ba}}^{\text{2+}}\text{(aq)}\to {\text{BaCrO}}_{\text{4}}(s)$

(b)

Interpretation Introduction

Interpretation:

Chemical equation, overall equation, complete ionic equation and also the net ionic equation for the reaction between ${\text{CuSO}}_{\text{4}}$ and ${\text{Na}}_{\text{2}}\text{S}$ has to be written.

Explanation of Solution

The reaction given is between aqueous solutions of ${\text{CuSO}}_{\text{4}}$ and ${\text{Na}}_{\text{2}}\text{S}$.  The reaction can be given as follows:

    ${\text{CuSO}}_{\text{4}}(\text{aq})+{\text{Na}}_{\text{2}}\text{S}(\text{aq})\to \text{CuS}(s)+{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq)}$

Aqueous solution of ${\text{CuSO}}_{\text{4}}$ reacts with ${\text{Na}}_{\text{2}}\text{S}$ resulting in formation of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ and $\text{CuS}$.  The precipitate that is formed is $\text{CuS}$.

The balanced chemical equation for the reaction is given as follows:

    ${\text{CuSO}}_{\text{4}}(\text{aq})+{\text{Na}}_{\text{2}}\text{S}(\text{aq})\to \text{CuS}(s)+{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq)}$

Complete ionic equation is written by showing the ions present in aqueous solution.  This is given as follows:

  ${\text{Cu}}^{\text{2+}}{\text{(aq)+SO}}_{\text{4}}{}^{2-}(\text{aq})+2{\text{Na}}^{\text{+}}{\text{(aq)+S}}^{2-}(\text{aq})\to \text{CuS}(s)+2{\text{Na}}^{\text{+}}{\text{(aq)+SO}}_{\text{4}}{}^{2-}(\text{aq})$

In the complete ionic equation, the common ions that are present on both sides of the equation is known as spectator ions.  Thus in this case, the spectator ions are ${\text{Na}}^{\text{+}}$ and ${\text{SO}}_{\text{4}}{}^{2-}$.  Net ionic equation can be obtained from the complete ionic equation by removing the spectator ions.

${\text{Cu}}^{\text{2+}}\text{(aq)+}\cancel{{\text{SO}}_{\text{4}}{}^{2-}(\text{aq})}+\cancel{2{\text{Na}}^{\text{+}}\text{(aq)}}{\text{+S}}^{2-}(\text{aq})\to \text{CuS}(s)+\cancel{2{\text{Na}}^{\text{+}}\text{(aq)}}\text{+}\cancel{{\text{SO}}_{\text{4}}{}^{2-}(\text{aq})}$

Thus net ionic equation can be written as follows:

    ${\text{Cu}}^{\text{2+}}{\text{(aq)+S}}^{2-}(\text{aq})\to \text{CuS}(s)$

(c)

Interpretation Introduction

Interpretation:

Chemical equation, overall equation, complete ionic equation and also the net ionic equation for the reaction between ${\text{FeCl}}_{\text{2}}$ and ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{3}}{\text{PO}}_{\text{4}}$ has to be written.

Explanation of Solution

The reaction given is between aqueous solutions of ${\text{FeCl}}_{\text{2}}$ and ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{3}}{\text{PO}}_{\text{4}}$.  The reaction can be given as follows:

    ${\text{3FeCl}}_{\text{2}}(\text{aq})+2{{\text{(NH}}_{\text{4}}\text{)}}_{\text{3}}{\text{PO}}_{\text{4}}(\text{aq})\to {\text{Fe}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}(s)+6{\text{NH}}_{\text{4}}\text{Cl(aq)}$

Aqueous solution of ${\text{FeCl}}_{\text{2}}$ reacts with ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{3}}{\text{PO}}_{\text{4}}$ resulting in formation of ${\text{Fe}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$ and ${\text{NH}}_{\text{4}}\text{Cl}$.  The precipitate that is formed is ${\text{Fe}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$.

The balanced chemical equation for the reaction is given as follows:

    ${\text{3FeCl}}_{\text{2}}(\text{aq})+2{{\text{(NH}}_{\text{4}}\text{)}}_{\text{3}}{\text{PO}}_{\text{4}}(\text{aq})\to {\text{Fe}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}(s)+6{\text{NH}}_{\text{4}}\text{Cl(aq)}$

Complete ionic equation is written by showing the ions present in aqueous solution.  This is given as follows:

${\text{3Fe}}^{\text{2+}}{\text{(aq)+6Cl}}^{-}(\text{aq})+6{\text{NH}}_{\text{4}}{}^{\text{+}}{\text{(aq)+2PO}}_{\text{4}}{}^{3-}(\text{aq})\to {\text{Fe}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}(s)+6{\text{NH}}_{\text{4}}{}^{\text{+}}{\text{(aq)+6Cl}}^{-}(\text{aq})$

In the complete ionic equation, the common ions that are present on both sides of the equation is known as spectator ions.  Thus in this case, the spectator ions are ${\text{NH}}_{\text{4}}{}^{\text{+}}$ and ${\text{Cl}}^{-}$.  Net ionic equation can be obtained from the complete ionic equation by removing the spectator ions.

${\text{3Fe}}^{\text{2+}}\text{(aq)+}\cancel{{\text{6Cl}}^{-}(\text{aq})}+\cancel{6{\text{NH}}_{\text{4}}{}^{\text{+}}\text{(aq)}}{\text{+2PO}}_{\text{4}}{}^{3-}(\text{aq})\to {\text{Fe}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}(s)+\cancel{6{\text{NH}}_{\text{4}}{}^{\text{+}}\text{(aq)}}\text{+}\cancel{{\text{6Cl}}^{-}(\text{aq})}$

Thus net ionic equation can be written as follows:

    ${\text{3Fe}}^{\text{2+}}{\text{(aq)+2PO}}_{\text{4}}{}^{3-}(\text{aq})\to {\text{Fe}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}(s)$

(d)

Interpretation Introduction

Interpretation:

Chemical equation, overall equation, complete ionic equation and also the net ionic equation for the reaction between ${\text{K}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}$ and ${\text{Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ has to be written.

Explanation of Solution

The reaction given is between aqueous solutions of ${\text{K}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}$ and ${\text{Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}$.  The reaction can be given as follows:

    ${\text{K}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}(\text{aq})+{\text{Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}(\text{aq})\to {\text{CaC}}_{\text{2}}{\text{O}}_{\text{4}}(s)+{\text{2KNO}}_{\text{3}}\text{(aq)}$

Aqueous solution of ${\text{K}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}$ reacts with ${\text{Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ resulting in formation of ${\text{CaC}}_{\text{2}}{\text{O}}_{\text{4}}$ and ${\text{KNO}}_{\text{3}}$.  The precipitate that is formed is ${\text{CaC}}_{\text{2}}{\text{O}}_{\text{4}}$.

The balanced chemical equation for the reaction is given as follows:

    ${\text{K}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}(\text{aq})+{\text{Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}(\text{aq})\to {\text{CaC}}_{\text{2}}{\text{O}}_{\text{4}}(s)+{\text{2KNO}}_{\text{3}}\text{(aq)}$

Complete ionic equation is written by showing the ions present in aqueous solution.  This is given as follows:

${\text{2K}}^{\text{+}}{\text{(aq)+C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{2-}(\text{aq})+{\text{Ca}}^{\text{2+}}{\text{(aq)+2NO}}_{\text{3}}{}^{-}(\text{aq})\to {\text{CaC}}_{\text{2}}{\text{O}}_{\text{4}}(s)+{\text{2K}}^{\text{+}}{\text{(aq)+2NO}}_{\text{3}}{}^{-}(\text{aq})$

In the complete ionic equation, the common ions that are present on both sides of the equation is known as spectator ions.  Thus in this case, the spectator ions are ${\text{K}}^{\text{+}}$ and ${\text{NO}}_{\text{3}}{}^{-}$.  Net ionic equation can be obtained from the complete ionic equation by removing the spectator ions.

$\cancel{{\text{2K}}^{\text{+}}\text{(aq)}}{\text{+C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{2-}(\text{aq})+{\text{Ca}}^{\text{2+}}\text{(aq)+}\cancel{{\text{2NO}}_{\text{3}}{}^{-}(\text{aq})}\to {\text{CaC}}_{\text{2}}{\text{O}}_{\text{4}}(s)+\cancel{{\text{2K}}^{\text{+}}\text{(aq)}}\text{+}\cancel{{\text{2NO}}_{\text{3}}{}^{-}(\text{aq})}$

Thus net ionic equation can be written as follows:

    ${\text{C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{2-}(\text{aq})+{\text{Ca}}^{\text{2+}}\text{(aq)}\to {\text{CaC}}_{\text{2}}{\text{O}}_{\text{4}}(s)$

(e)

Interpretation Introduction

Interpretation:

Chemical equation, overall equation, complete ionic equation and also the net ionic equation for the reaction between ${\text{NiSO}}_{\text{4}}$ and ${\text{Ba(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ has to be written.

Explanation of Solution

The reaction given is between aqueous solutions of ${\text{NiSO}}_{\text{4}}$ and ${\text{Ba(NO}}_{\text{3}}{\text{)}}_{\text{2}}$.  The reaction can be given as follows:

    ${\text{NiSO}}_{\text{4}}(\text{aq})+{\text{Ba(NO}}_{\text{3}}{\text{)}}_{\text{2}}(\text{aq})\to {\text{BaSO}}_{\text{4}}(s)+{\text{Ni(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{(aq)}$

Aqueous solution of ${\text{NiSO}}_{\text{4}}$ reacts with ${\text{Ba(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ resulting in formation of ${\text{BaSO}}_{\text{4}}$ and ${\text{Ni(NO}}_{\text{3}}{\text{)}}_{\text{2}}$.  The precipitate that is formed is ${\text{BaSO}}_{\text{4}}$.

The balanced chemical equation for the reaction is given as follows:

    ${\text{NiSO}}_{\text{4}}(\text{aq})+{\text{Ba(NO}}_{\text{3}}{\text{)}}_{\text{2}}(\text{aq})\to {\text{BaSO}}_{\text{4}}(s)+{\text{Ni(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{(aq)}$

Complete ionic equation is written by showing the ions present in aqueous solution.  This is given as follows:

  ${\text{Ni}}^{\text{2+}}{\text{(aq)+SO}}_{\text{4}}{}^{2-}(\text{aq})+{\text{Ba}}^{\text{2+}}{\text{(aq)+2NO}}_{\text{3}}{}^{-}(\text{aq})\to {\text{BaSO}}_{\text{4}}(s)+{\text{Ni}}^{\text{2+}}{\text{(aq)+2NO}}_{\text{3}}{}^{-}(\text{aq})$

In the complete ionic equation, the common ions that are present on both sides of the equation is known as spectator ions.  Thus in this case, the spectator ions are ${\text{Ni}}^{\text{2+}}$ and ${\text{NO}}_{\text{3}}{}^{-}$.  Net ionic equation can be obtained from the complete ionic equation by removing the spectator ions.

$\cancel{{\text{Ni}}^{\text{2+}}\text{(aq)}}{\text{+SO}}_{\text{4}}{}^{2-}(\text{aq})+{\text{Ba}}^{\text{2+}}\text{(aq)+}\cancel{{\text{2NO}}_{\text{3}}{}^{-}(\text{aq})}\to {\text{BaSO}}_{\text{4}}(s)+\cancel{{\text{Ni}}^{\text{2+}}\text{(aq)}}\text{+}\cancel{{\text{2NO}}_{\text{3}}{}^{-}(\text{aq})}$

Thus net ionic equation can be written as follows:

    ${\text{SO}}_{\text{4}}{}^{2-}(\text{aq})+{\text{Ba}}^{\text{2+}}\text{(aq)}\to {\text{BaSO}}_{\text{4}}(s)$