8. a solution of ammonium sulfate is added to a saturated solution of barium hydroxide.\nmolecular…

Explanation:

Step1: Write molecular equation

$$(NH_4)_2SO_4(aq)+Ba(OH)_2(aq)\rightarrow BaSO_4(s)+2NH_3(g)+2H_2O(l)$$

Step2: Write complete ionic equation

Break all soluble ionic compounds into ions. $(NH_4)_2SO_4$ breaks into $2NH_4^+(aq)+SO_4^{2 - }(aq)$, $Ba(OH)_2$ breaks into $Ba^{2+}(aq)+2OH^-(aq)$, and since $BaSO_4$ is insoluble, $NH_3$ is a gas and $H_2O$ is a liquid, they remain in their molecular forms. So the complete - ionic equation is $2NH_4^+(aq)+SO_4^{2 - }(aq)+Ba^{2+}(aq)+2OH^-(aq)\rightarrow BaSO_4(s)+2NH_3(g)+2H_2O(l)$

Step3: Write net - ionic equation

Spectator ions are ions that appear on both sides of the complete ionic equation. Here, the spectator ions are $NH_4^+$ and $OH^-$. After removing them, the net - ionic equation is $SO_4^{2 - }(aq)+Ba^{2+}(aq)\rightarrow BaSO_4(s)$

Step4: Identify spectator ions

The spectator ions are $NH_4^+(aq)$ and $OH^-(aq)$

Answer:

Molecular: $(NH_4)_2SO_4(aq)+Ba(OH)_2(aq)\rightarrow BaSO_4(s)+2NH_3(g)+2H_2O(l)$ Complete Ionic: $2NH_4^+(aq)+SO_4^{2 - }(aq)+Ba^{2+}(aq)+2OH^-(aq)\rightarrow BaSO_4(s)+2NH_3(g)+2H_2O(l)$ Net Ionic: $SO_4^{2 - }(aq)+Ba^{2+}(aq)\rightarrow BaSO_4(s)$ Spectator ions: $NH_4^+(aq), OH^-(aq)$