Write ionic and net ionic equations for the following reactions: (a) \(\mathrm{Na}_{2} \mathrm{~S}(a q)+\mathrm{ZnCl}_{2}(a q) \longrightarrow\) (b) \(\mathrm{K}_{3} \mathrm{PO}_{4}(a q)+3 \mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}(a q) \longrightarrow\) (c) \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{NaOH}(a q) \longrightarrow\)

Short Answer

Expert verified
a) Net Ionic Equation : \(\mathrm{S^{2-}(aq) + Zn^{2+}(aq)} \longrightarrow\)b) Net Ionic Equation : \(\mathrm{PO_{4}^{3-}(aq) + 3Sr^{2+}(aq)} \longrightarrow\)c) Net Ionic Equation : \(\mathrm{Mg^{2+}(aq) + 2OH^{-}(aq)} \longrightarrow\)

Step by step solution

01

Ionic and Net Ionic Equations

Ionic Equations are equations that show all the compounds in ionic form. Net Ionic Equations are equations that only show the key substances involved in the reaction. A rule to remember, all soluble ionic compounds and strong acids dissociate into their ions in solution.
02

Problem a: \(\mathrm{Na}_{2} \mathrm{~S}(a q)+\mathrm{ZnCl}_{2}(a q) \longrightarrow\)

First, break down the reactants in ionic form and this is the full ionic equation:\(\mathrm{2Na^{+}(aq) + S^{2-}(aq) + Zn^{2+}(aq) + 2Cl^{-}(aq)} \longrightarrow\)Next, the net ionic equation is generated by removing spectator ions, ions which occur in the same form on both sides of the equation. These are \(Na^{+}\) and \(Cl^{-}\).Hence, the net ionic equation is \(\mathrm{S^{2-}(aq) + Zn^{2+}(aq)} \longrightarrow\)
03

Problem b: \(\mathrm{K}_{3} \mathrm{PO}_{4}(a q)+3 \mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}(a q) \longrightarrow\)

Start with the full ionic equation:\(\mathrm{3K^{+}(aq) + PO_{4}^{3-}(aq) + 3Sr^{2+}(aq) + 6NO_{3}^{-}(aq)} \longrightarrow\)Remove the spectator ions, which are \(K^{+}\) and \(NO_{3}^{-}\), to get the net ionic equation:\(\mathrm{PO_{4}^{3-}(aq) + 3Sr^{2+}(aq)} \longrightarrow\)
04

Problem c: \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{NaOH}(a q) \longrightarrow\)

Break down the reactants into ions for the full ionic equation:\(\mathrm{Mg^{2+}(aq) + 2NO_{3}^{-}(aq) + 2Na^{+}(aq) + 2OH^{-}(aq)} \longrightarrow\)Remove the spectator ions, here again, \(Na^{+}\) and \(NO_{3}^{-}\), to find the net ionic equation:\(\mathrm{Mg^{2+}(aq) + 2OH^{-}(aq)} \longrightarrow\)

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