Explanation:
The formula for Mohr's salt is
\(\mathrm{FeSO}_{4} \cdot\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}\).
\[\begin{array}{l}
{\rm{FeS}}{{\rm{O}}_4} \cdot {\left( {{\rm{N}}{{\rm{H}}_{\rm{4}}}} \right)_{\rm{2}}}{\rm{S}}{{\rm{O}}_4} \cdot {\rm{6}}{{\rm{H}}_{\rm{2}}}{\rm{O}} \to {\rm{F}}{{\rm{e}}^{{\rm{2 + }}}}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{ + 2NH}}_{\rm{4}}^{\rm{ + }}\,{\rm{ + }}\,{\rm{2SO}}_{\rm{4}}^{{\rm{2}} - }\,{\rm{ + }}\,{\rm{6}}{{\rm{H}}_{\rm{2}}}{\rm{O}}
\end{array}\]
Therefore, three cations are produced by Mohr's salt in the solution.