Question

Complete the following precipitation reactions with balanced molecular, total ionic, and net ionic equations:

$\begin{array}{l}\mathbf{\left(}\mathbf{a}\mathbf{\right)}{\mathbf{Hg}}_{\mathbf{2}}{\mathbf{\left(}{\mathbf{NO}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{KI}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\to }\\ \mathbf{\left(}\mathbf{b}\mathbf{\right)}{\mathbf{FeSO}}_{\mathbf{4}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{Sr}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{}\mathbf{\to }\end{array}$

Short answer

Precipitation of ions depends upon the charge of ions present in a given solution.

Step-by-step solution

Write the equation of  Hg2(NO3)2and KI

The molecular reaction is:

${\mathrm{Hg}}_2{\left({\mathrm{NO}}_3\right)}_2\left(\mathrm{aq}\right)+\mathrm{KI}\left(\mathrm{aq}\right)\to {\mathrm{Hg}}_2{\mathrm{I}}_2\left(\mathrm{s}\right)+2{\mathrm{KNO}}_3\left(\mathrm{aq}\right)$

The total ionic equation is:

$2{\mathrm{Hg}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{NO}}_3^{-}\left(\mathrm{aq}\right)+2{\mathrm{K}}^{+}\left(\mathrm{aq}\right)+2{\mathrm{I}}^{-}\to 2{\mathrm{K}}^{+}\left(\mathrm{aq}\right)+2{\mathrm{NO}}_3^{-}\left(\mathrm{aq}\right)+{\mathrm{Hg}}_2{\mathrm{l}}_2\left(\mathrm{s}\right)$

The net ionic equation is:

$2{\mathrm{Hg}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{I}}^{-}\left(\mathrm{aq}\right)\to {\mathrm{Hg}}_2{\mathrm{l}}_2\left(\mathrm{s}\right)$

Write the equation of FeSO4(aq)  and Sr (OH)2

The molecular reaction is:

${\mathrm{FeSO}}_4\left(\mathrm{aq}\right)+\mathrm{Sr}{\left(\mathrm{OH}\right)}_2\left(\mathrm{aq}\right)\to \mathrm{Fe}{\left(\mathrm{OH}\right)}_2\left(\mathrm{s}\right)+{\mathrm{SrSO}}_4\left(\mathrm{s}\right)$

The total ionic equation is:

${\mathrm{Fe}}^{+2}\left(\mathrm{aq}\right)+{\mathrm{SO}}_4\left(\mathrm{aq}\right)+{\mathrm{Sr}}^{+2}\left(\mathrm{aq}\right)+2{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)\to \mathrm{Fe}{\left(\mathrm{OH}\right)}_2\left(\mathrm{s}\right)+{\mathrm{SrSO}}_4\left(\mathrm{s}\right)$

The net ionic equation is:

${\mathrm{Fe}}^{+2}\left(\mathrm{aq}\right)+{\mathrm{SO}}_4\left(\mathrm{aq}\right)+{\mathrm{Sr}}^{+2}\left(\mathrm{aq}\right)+2{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)\to \mathrm{Fe}{\left(\mathrm{OH}\right)}_2\left(\mathrm{s}\right)+{\mathrm{SrSO}}_4\left(\mathrm{s}\right)$

Since this reaction does not have any spectator ion, hence both ionic and net ionic reaction is the same