Chapter 15, Problem 29PE

(a)

Interpretation Introduction

Interpretation:

The indicated equation has to be balanced and then converted to net ionic equation.

  ${\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}\left(aq\right)+\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\left(aq\right)\to \text{Ca}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(s\right)+{\text{KNO}}_{\text{3}}\left(aq\right)$

Concept Introduction:

The ionic equation can be written based on the rules as follows:

• Strong electrolytes dissociate completely into ionic forms.
• Weak electrolytes remain in molecular form.
• Nonelectrolytes remain in molecular form.
• Insoluble precipitates and gases remain in molecular forms.
• The net ionic equation includes only ions that have reacted and thus any spectator ions are usually omitted.

Explanation of Solution

The unbalanced equation is written as follows:

  ${\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}\left(aq\right)+\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\left(aq\right)\to {\text{Ca}}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(s\right)+{\text{KNO}}_{\text{3}}\left(aq\right)$        (1)

Since there are two ${\text{PO}}_{\text{4}}$ units in ${\text{Ca}}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ while only 1 in ${\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}$, so place 2 as coefficient for ${\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}$ in equation (1).

  ${\text{2K}}_{\text{3}}{\text{PO}}_{\text{4}}\left(aq\right)+\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\left(aq\right)\to {\text{Ca}}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(s\right)+{\text{KNO}}_{\text{3}}\left(aq\right)$        (2)

Since there are three $\text{Ca}$ units in ${\text{Ca}}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ while only 1 in $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$, so place 2 as coefficient for ${\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}$ and 6 as coefficient of ${\text{KNO}}_{\text{3}}$ in equation (2) to balance $\text{K}$ atoms. Therefore complete balanced reaction can be written as follows:

  ${\text{2K}}_{\text{3}}{\text{PO}}_{\text{4}}\left(aq\right)+3\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\left(aq\right)\to {\text{Ca}}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(s\right)+6{\text{KNO}}_{\text{3}}\left(aq\right)$

Since ${\text{Ca}}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is solid so total ionic equation with all the ions in dissociated form gives total equation as follows:

  $\begin{array}{c}{\text{6K}}^{+}\left(aq\right)+{\text{2PO}}_4^{3-}\left(aq\right)+3{\text{Ca}}^{2+}\left(aq\right)+6{\text{NO}}_3^{-}\left(aq\right)\to {\text{Ca}}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(s\right)+\\ 6{\text{K}}^{+}\left(aq\right)+6{\text{NO}}_3^{-}\left(aq\right)\end{array}$

The common ${\text{NO}}_3^{-}$ and ${\text{K}}^{+}$ ions are cancelled out from each side to obtain final net ionic equation as follows:

  ${\text{2PO}}_4^{3-}\left(aq\right)+3{\text{Ca}}^{2+}\left(aq\right)\to {\text{Ca}}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(s\right)$

(b)

Interpretation Introduction

Interpretation:

The indicated equation has to be balanced and then converted to net ionic equation.

  $\text{Al}\left(s\right)+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\to {\text{H}}_{\text{2}}\left(g\right)+{\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_3\left(aq\right)$

Concept Introduction:

Refer to part (a).

Explanation of Solution

The unbalanced equation is written as follows:

  $\text{Al}\left(s\right)+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\to {\text{H}}_{\text{2}}\left(g\right)+{\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_3\left(aq\right)$        (3)

Since there are three ${\text{SO}}_{\text{4}}$ units in ${\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_3$ while only 1 in ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$, so place 3 as coefficient for ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ in equation (3).

  $\text{2Al}\left(s\right)+3{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\to 3{\text{H}}_{\text{2}}\left(g\right)+{\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_3\left(aq\right)$        (4)

Since there are two $\text{Al}$ units in ${\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_3$ while only 1 on left so place 2 as coefficient for $\text{Al}$ and similarly place 3 as coefficient of ${\text{H}}_{\text{2}}$ to balance $\text{H}$ atoms in equation (4).  Therefore complete balanced reaction can be written as follows:

  $\text{2Al}\left(s\right)+3{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\to 3{\text{H}}_{\text{2}}\left(g\right)+{\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_3\left(aq\right)$

Since $\text{Al}$ is solid and ${\text{H}}_{\text{2}}$ is gas so total ionic equation with all the ions in dissociated form gives total equation as follows:

  $\text{2Al}\left(s\right)+6{\text{H}}^{+}\left(aq\right)+3{\text{SO}}_4^{2-}\left(aq\right)\to 3{\text{H}}_{\text{2}}\left(g\right)+2{\text{Al}}^{3+}\left(aq\right)+3{\text{SO}}_4^{2-}\left(aq\right)$

The common ${\text{SO}}_4^{2-}$ ions are cancelled out from each side to obtain final net ionic equation as follows:

  $\text{2Al}\left(s\right)+6{\text{H}}^{+}\left(aq\right)\to 3{\text{H}}_{\text{2}}\left(g\right)+2{\text{Al}}^{3+}\left(aq\right)$

(c)

Interpretation Introduction

Interpretation:

The indicated equation has to be balanced and then converted to net ionic equation.

  ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(aq\right)+\text{HCl}\left(aq\right)\to \text{NaCl}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{CO}}_{\text{2}}\left(g\right)$

Concept Introduction:

Refer to part (a).

Explanation of Solution

The unbalanced equation is written as follows:

  ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(aq\right)+\text{HCl}\left(aq\right)\to \text{NaCl}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{CO}}_{\text{2}}\left(g\right)$        (5)

Since there are two $\text{Na}$ units in ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ while only 1 on left in $\text{NaCl}$ so place 2 as coefficient for $\text{NaCl}$ and similarly place 2 as coefficient of $\text{HCl}$ to balance $\text{Na}$ and $\text{H}$ atoms in equation (5).  Therefore complete balanced reaction can be written as follows:

  ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(aq\right)+2\text{HCl}\left(aq\right)\to 2\text{NaCl}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{CO}}_{\text{2}}\left(g\right)$

Since ${\text{H}}_{\text{2}}\text{O}$ is liquid and ${\text{CO}}_{\text{2}}$ is gas so total ionic equation with all ions in dissociated form gives total equation as follows:

  $\begin{array}{c}{\text{2Na}}^{+}\left(aq\right)+{\text{CO}}_3^{2-}\left(aq\right)+2{\text{H}}^{+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)\to 2{\text{Na}}^{+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)\\ +{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{CO}}_{\text{2}}\left(g\right)\end{array}$

The common ${\text{H}}^{+}$ and ${\text{Cl}}^{-}$ ions are cancelled out from each side to obtain final net ionic equation as follows:

  ${\text{CO}}_3^{2-}\left(aq\right)+2{\text{H}}^{+}\left(aq\right)\to {\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{CO}}_{\text{2}}\left(g\right)$