Question

Because theoxide ion is basic, metal oxides react readily with acids. (a) Write the net ionic equation for the following reaction: $$ \mathrm{FeO}(s)+2 \mathrm{HClO}_{4}(a q) \longrightarrow \mathrm{Fe}\left(\mathrm{ClO}_{4}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ (b) Based on the equation in part (a), write the net ionic equation for the reaction that occurs between \(\mathrm{NiO}(s)\) and an aqueous solution of nitric acid.

Short answer

(a) The net ionic equation for the reaction between FeO and HClO4 is: \[ \mathrm{FeO}(s) + 2\mathrm{H}^{+}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q) + \mathrm{H}_{2}\mathrm{O}(l) \] (b) The net ionic equation for the reaction between NiO and an aqueous solution of nitric acid is: \[ \mathrm{NiO}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Ni}^{2+}(a q) + \mathrm{H}_{2}\mathrm{O}(l) \]

Step-by-step solution

Write the balanced chemical equation for part (a)

The given balanced chemical equation is: $$ \mathrm{FeO}(s)+2 \mathrm{HClO}_{4}(a q) \longrightarrow \mathrm{Fe}\left(\mathrm{ClO}_{4}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$

Split the reactants and products into their respective ions for part (a)

Split the reactants and products that are in aqueous solution into their respective ions: $$ \mathrm{FeO}(s) + 2\mathrm{H}^{+}(a q) + 2\mathrm{ClO}_4^{-}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q) + 2\mathrm{ClO}_4^{-}(a q) + \mathrm{H}_{2}\mathrm{O}(l) $$ Since FeO is a solid and H2O is a liquid, we don't split them into ions.

Write the net ionic equation for part (a)

Now, remove the spectator ions from the equation (those that appear on both sides of the equation). In this case, we have the ClO4- ion as spectator ions: $$ \mathrm{FeO}(s) + 2\mathrm{H}^{+}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q) + \mathrm{H}_{2}\mathrm{O}(l) $$ This is the net ionic equation for part (a).

Write the balanced chemical equation for part (b)

Based on the equation from part (a), we are asked to write the net ionic equation for the reaction between NiO and an aqueous solution of nitric acid (HNO3). First, write the balanced chemical equation for this reaction: $$ \mathrm{NiO}(s) + 2 \mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q) + \mathrm{H}_{2}\mathrm{O}(l) $$

Split the reactants and products into their respective ions for part (b)

Now, split the reactants and products that are in aqueous solution into their respective ions: $$ \mathrm{NiO}(s) + 2\mathrm{H}^{+}(a q) + 2\mathrm{NO}_3^{-}(a q) \longrightarrow \mathrm{Ni}^{2+}(a q) + 2\mathrm{NO}_3^{-}(a q) + \mathrm{H}_{2}\mathrm{O}(l) $$ Again, we don't split NiO and H2O, because they are a solid and a liquid respectively.

Write the net ionic equation for part (b)

Finally, remove the spectator ions from the equation (those that appear on both sides of the equation). In this case, we have the NO3- ion as spectator ions: $$ \mathrm{NiO}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Ni}^{2+}(a q) + \mathrm{H}_{2}\mathrm{O}(l) $$ This is the net ionic equation for part (b).