Chapter 15, Problem 6PE

(a)

Interpretation Introduction

Interpretation:

Total and net ionic equation for reaction indicated has to be written.

  $\text{Mg}\left(s\right)+2{\text{HClO}}_4\left(aq\right)\to \text{Mg}{\left({\text{ClO}}_4\right)}_2\left(aq\right)+{\text{H}}_{\text{2}}\left(g\right)$

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 formula equation is given as follows:

  $\text{Mg}\left(s\right)+2{\text{HClO}}_4\left(aq\right)\to \text{Mg}{\left({\text{ClO}}_4\right)}_2\left(aq\right)+{\text{H}}_{\text{2}}\left(g\right)$

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

  $\text{Mg}\left(s\right)+2{\text{H}}^{+}\left(aq\right)+{\text{2ClO}}_4^{-}\left(aq\right)\to {\text{Mg}}^{2+}\left(aq\right)+{\text{2ClO}}_4^{-}\left(aq\right)+{\text{H}}_{\text{2}}\left(g\right)$

The common ${\text{ClO}}_4^{-}$ ions are cancelled out from each side and final net ionic equation is written as follows:

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

(b)

Interpretation Introduction

Interpretation:

Total and net ionic equation for reaction indicated has to be written.

  $\text{2Al}\left(s\right)+3{\text{H}}_2{\text{SO}}_4\left(aq\right)\to {\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_3\left(aq\right)+3{\text{H}}_{\text{2}}\left(g\right)$

Concept Introduction:

Refer to part (a).

Explanation of Solution

The formula equation is given as follows:

  $\text{2Al}\left(s\right)+3{\text{H}}_2{\text{SO}}_4\left(aq\right)\to {\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_3\left(aq\right)+3{\text{H}}_{\text{2}}\left(g\right)$

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

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

The common ${\text{SO}}_4^{2-}$ ions are cancelled out from each side and thus net ionic equation is written as follows:

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

(c)

Interpretation Introduction

Interpretation:

Total and net ionic equation for reaction indicated has to be written.

  $\text{2NaOH}\left(s\right)+{\text{H}}_2{\text{CO}}_{\text{3}}\left(aq\right)\to {\text{Na}}_2{\text{CO}}_{\text{3}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Concept Introduction:

Refer to part (a).

Explanation of Solution

The formula equation is given as follows:

  $\text{2NaOH}\left(s\right)+{\text{H}}_2{\text{CO}}_{\text{3}}\left(aq\right)\to {\text{Na}}_2{\text{CO}}_{\text{3}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Since ${\text{H}}_{\text{2}}\text{O}$ is liquid and $\text{NaOH}$ is solid so total ionic equation with only ions in dissociated form gives total equation written as follows:

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

The common ${\text{CO}}_3^{2-}$ ions are cancelled out from each side and thus net ionic equation is written as follows:

  $\text{2NaOH}\left(s\right)+2{\text{H}}^{+}\left(aq\right)\to 2{\text{Na}}^{+}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

(d)

Interpretation Introduction

Interpretation:

Total and net ionic equations for reaction indicated have to be written.

  $\text{Ba}{\left(\text{OH}\right)}_2\left(s\right)+2{\text{HClO}}_{\text{4}}\left(aq\right)\to \text{Ba}{\left({\text{ClO}}_{\text{4}}\right)}_2\left(aq\right)+2{\text{H}}_2\text{O}\left(l\right)$

Concept Introduction:

Refer to part (a).

Explanation of Solution

The formula equation is given as follows:

  $\text{Ba}{\left(\text{OH}\right)}_2\left(s\right)+2{\text{HClO}}_{\text{4}}\left(aq\right)\to \text{Ba}{\left({\text{ClO}}_{\text{4}}\right)}_2\left(aq\right)+2{\text{H}}_2\text{O}\left(l\right)$

Since ${\text{H}}_{\text{2}}\text{O}$ is liquid and $\text{Ba}{\left(\text{OH}\right)}_2$ is solid so total ionic equation with only ions in dissociated form is written as follows:

  $\text{Ba}{\left(\text{OH}\right)}_2\left(s\right)+3{\text{H}}^{+}\left(aq\right)+2{\text{ClO}}_4^{-}\left(aq\right)\to {\text{Ba}}^{2+}\left(aq\right)+2{\text{ClO}}_4^{-}\left(aq\right)+2{\text{H}}_2\text{O}\left(l\right)$

The common $2{\text{ClO}}_4^{-}$ ions are cancelled out from each side and thus net ionic equation is written as follows:

  $\text{Ba}{\left(\text{OH}\right)}_2\left(s\right)+3{\text{H}}^{+}\left(aq\right)\to {\text{Ba}}^{2+}\left(aq\right)+2{\text{H}}_2\text{O}\left(l\right)$