11. a solution of barium chloride is added drop by drop to a solution of sodium carbonate, causing a…

Explanation:

Step1: Write molecular equation products

The reaction between $BaCl_2$ and $Na_2CO_3$ forms $BaCO_3$ (precipitate) and $NaCl$. So the molecular equation is $BaCl_2(aq)+Na_2CO_3(aq)\rightarrow BaCO_3(s)+2NaCl(aq)$.

Step2: Write complete - ionic equation

Ionize aqueous compounds. $BaCl_2$ ionizes to $Ba^{2 +}(aq)+2Cl^{-}(aq)$, $Na_2CO_3$ ionizes to $2Na^{+}(aq)+CO_3^{2 -}(aq)$, and $NaCl$ ionizes to $2Na^{+}(aq)+2Cl^{-}(aq)$. So the complete - ionic equation is $Ba^{2 +}(aq)+2Cl^{-}(aq)+2Na^{+}(aq)+CO_3^{2 -}(aq)\rightarrow BaCO_3(s)+2Na^{+}(aq)+2Cl^{-}(aq)$.

Step3: Write net - ionic equation

Cancel out spectator ions ($Na^{+}$ and $Cl^{-}$). The net - ionic equation is $Ba^{2 +}(aq)+CO_3^{2 -}(aq)\rightarrow BaCO_3(s)$.

Step4: Identify spectator ions

Spectator ions are those that do not participate in the reaction. Here, $Na^{+}$ and $Cl^{-}$ are spectator ions.

Answer:

Molecular: $BaCl_2(aq)+Na_2CO_3(aq)\rightarrow BaCO_3(s)+2NaCl(aq)$ Complete Ionic: $Ba^{2 +}(aq)+2Cl^{-}(aq)+2Na^{+}(aq)+CO_3^{2 -}(aq)\rightarrow BaCO_3(s)+2Na^{+}(aq)+2Cl^{-}(aq)$ Net Ionic: $Ba^{2 +}(aq)+CO_3^{2 -}(aq)\rightarrow BaCO_3(s)$ Spectator ions: $Na^{+}(aq), Cl^{-}(aq)$