Question

Complete and balance the following molecular equations, and then write the net ionic equation for each: (a) \(\mathrm{HBr}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q) \longrightarrow\) (b) \(\mathrm{Cu}(\mathrm{OH})_{2}(s)+\mathrm{HClO}_{4}(a q) \longrightarrow\) (c) \(\mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{HNO}_{3}(a q) \longrightarrow\)

Short answer

The net ionic equations for the given reactions are: (a) \(2\mathrm{H^{+}}(aq) + 2\mathrm{OH^{-}}(aq) \longrightarrow 2\mathrm{H_{2}O}(l)\) (b) \(\mathrm{Cu(OH)_{2}}(s) + 2\mathrm{H^{+}}(aq) \longrightarrow \mathrm{Cu^{2+}}(aq) + 2\mathrm{H_{2}O}(l)\) (c) \(\mathrm{Al(OH)_{3}}(s) + 3\mathrm{H^{+}}(aq) \longrightarrow \mathrm{Al^{3+}}(aq) + 3\mathrm{H_{2}O}(l)\)

Step-by-step solution

Reaction (a) - Write unbalanced molecular equation

The given unbalanced equation is: HBr(aq) + Ca(OH)2(aq) ->

Reaction (a) - Balance the equation

Let's balance the equation. We have \(2 \mathrm{HBr} (a q) + \mathrm{Ca}(\mathrm{OH})_{2}(a q) \longrightarrow \mathrm{CaBr}_{2} (\mathrm{sq}) + 2\mathrm{H}_{2}\mathrm{O}(\ell)\)

Reaction (a) - Complete ionic equation

Now write the complete ionic equation: \(2\mathrm{H^{+}}(aq) + 2\mathrm{Br^{-}}(aq) + \mathrm{Ca^{2+}}(aq) + 2\mathrm{OH^{-}}(aq) \longrightarrow \mathrm{Ca^{2+}}(aq) + 2\mathrm{Br^{-}}(aq) + 2\mathrm{H_{2}O}(l)\)

Reaction (a) - Net ionic equation

Finally, the net ionic equation is: \(2\mathrm{H^{+}}(aq) + 2\mathrm{OH^{-}}(aq) \longrightarrow 2\mathrm{H_{2}O}(l)\) For Reaction (b):

Reaction (b) - Write unbalanced molecular equation

The given unbalanced equation is: Cu(OH)2(s) + HClO4(aq) ->

Reaction (b) - Balance the equation

Let's balance the equation. We have \( \mathrm{Cu}(\mathrm{OH})_{2}(s) + 2\mathrm{HClO}_4(\mathrm{aq}) \longrightarrow \mathrm{Cu}(\mathrm{ClO}_{4})_{2}(\mathrm{sq}) + 2\mathrm{H}_{2}\mathrm{O}(\ell)\)

Reaction (b) - Complete ionic equation

Now write the complete ionic equation: \(\mathrm{Cu(OH)_{2}}(s) + 2\mathrm{H^{+}}(aq) + 2\mathrm{ClO}_{4}^-(aq) \longrightarrow \mathrm{Cu^{2+}}(aq) + 2\mathrm{ClO}_{4}^-(aq) + 2\mathrm{H_{2}O}(l)\)

Reaction (b) - Net ionic equation

Finally, the net ionic equation is: \(\mathrm{Cu(OH)_{2}}(s) + 2\mathrm{H^{+}}(aq) \longrightarrow \mathrm{Cu^{2+}}(aq) + 2\mathrm{H_{2}O}(l)\) For Reaction (c):

Reaction (c) - Write unbalanced molecular equation

The given unbalanced equation is: Al(OH)3(s) + HNO3(aq) ->

Reaction (c) - Balance the equation

Let's balance the equation. We have \(\mathrm{Al}(\mathrm{OH})_{3}(s) + 3\mathrm{HNO}_3(\mathrm{aq}) \longrightarrow \mathrm{Al}(\mathrm{NO}_{3})_{3}(\mathrm{sq}) + 3\mathrm{H}_{2}\mathrm{O}(\ell)\)

Reaction (c) - Complete ionic equation

Now write the complete ionic equation: \(\mathrm{Al(OH)_{3}}(s) + 3\mathrm{H^{+}}(aq) + 3\mathrm{NO}_{3}^-(aq) \longrightarrow \mathrm{Al^{3+}}(aq) + 3\mathrm{NO}_{3}^-(aq) + 3\mathrm{H_{2}O}(l)\)

Reaction (c) - Net ionic equation

Finally, the net ionic equation is: \(\mathrm{Al(OH)_{3}}(s) + 3\mathrm{H^{+}}(aq) \longrightarrow \mathrm{Al^{3+}}(aq) + 3\mathrm{H_{2}O}(l)\)