Complete each of the following as a net ionic equation. If no reaction occurs, so state. (a) $\mathrm{Ca}^{2+}+2 \mathrm{I}^{-}+2 \mathrm{Na}^{+}+\mathrm{CO}_{3}^{2-} \longrightarrow$ (b) $\mathrm{Ba}^{2+}+\mathrm{S}^{2-}+2 \mathrm{Na}^{+}+\mathrm{SO}_{4}^{2-} \longrightarrow$ (c) $2 \mathrm{K}^{+}+\mathrm{S}^{2-}+\mathrm{Ca}^{2+}+2 \mathrm{Cl}^{-} \longrightarrow$

Short Answer

Expert verified
(a) \(\mathrm{Ca}^{2+}+ \mathrm{CO}_{3}^{2-}\rightarrow \mathrm{CaCO}_3\) (precipitate), (b) \(\mathrm{Ba}^{2+} + \mathrm{SO}_{4}^{2-} \longrightarrow \mathrm{BaSO_4}\) (precipitate), (c) No reaction.

Step by step solution

01

Identify the compounds

The compounds are formed by pairing a cation (+) with an anion (-). For (a) we have \(\mathrm{CaI}_2\) and \(\mathrm{Na}_2 \mathrm{CO}_3\), for (b) we have \(\mathrm{Na}_2 \mathrm{S}\) and \(\mathrm{BaSO}_4\) and for (c) we have \(\mathrm{K}_2 \mathrm{S}\) and \(\mathrm{CaCl}_2\).
02

Full ionic equations

(a) \(\mathrm{Ca}^{2+}+2 \mathrm{I}^{-}+2 \mathrm{Na}^{+}+\mathrm{CO}_{3}^{2-} \longrightarrow \mathrm{Ca^{2+}}+2\mathrm{I^{-}}+2\mathrm{Na^{+}} +\mathrm{CO_3^{2-}}\),\n(b) \(\mathrm{Ba}^{2+}+\mathrm{S}^{2-}+2 \mathrm{Na}^{+}+\mathrm{SO}_{4}^{2-} \longrightarrow \mathrm{Ba^{2+}}+\mathrm{SO_4^{2-}}+2\mathrm{Na^{+}}+\mathrm{S^{2-}}\),\n(c) \(2 \mathrm{K}^{+}+\mathrm{S}^{2-}+\mathrm{Ca}^{2+}+2 \mathrm{Cl}^{-}\longrightarrow 2\mathrm{K^{+}}+\mathrm{S^{2-}}+\mathrm{Ca^{2+}}+2\mathrm{Cl^{-}}\).
03

Identify and exclude the spectator ions

The ions that do not change from reactants to products are the spectator ions. Exclude them to get the net ionic equations.\n(a) \(\mathrm{Ca}^{2+} + \mathrm{CO}_{3}^{2-} \longrightarrow \mathrm{CaCO_3}\) (precipitate),\n(b) \(\mathrm{Ba}^{2+} + \mathrm{SO}_{4}^{2-} \longrightarrow \mathrm{BaSO_4}\) (precipitate),\n(c) No reaction.

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