Write the balanced net ionic equation for the reaction when aqueous $NH_4Cl$ and aqueous $LiOH$ are mixed.

Question

What are the product(s) that remain on the products side after the spectator ions have been canceled?

A. NH_3(g) + H_2O(\ell) + LiCl(s)

B. LiCl(aq) + NH^+ + OH^-(aq)

C. LiCl(s) + NH_4OH(aq)

D. NH_3(g) + H_2O(\ell)

E. There is no reaction.

Answer 1

To determine the balanced net ionic equation for the reaction between aqueous \( NH_4Cl \) and aqueous \( LiOH \) and identify the product(s) that remain after canceling spectator ions, we can start by writing the chemical equation:

1. The dissociation of \( NH_4Cl \) and \( LiOH \) in water is as follows:

   • \( NH_4Cl \) dissociates into \( NH_4^+ \) and \( Cl^- \): \[ NH_4Cl(aq) \rightarrow NH_4^+(aq) + Cl^-(aq) \]
   • \( LiOH \) dissociates into \( Li^+ \) and \( OH^- \): \[ LiOH(aq) \rightarrow Li^+(aq) + OH^-(aq) \]

2. The complete ionic equation for the reaction can be written as: \[ NH_4^+(aq) + Cl^-(aq) + Li^+(aq) + OH^-(aq) \rightarrow ? \]

3. When \( NH_4^+ \) reacts with \( OH^- \), it produces \( NH_3 \) and \( H_2O \): \[ NH_4^+(aq) + OH^-(aq) \rightarrow NH_3(g) + H_2O(\ell) \]

4. Incorporating the spectator ions, \( Cl^- \) and \( Li^+ \), we can write the complete ionic equation: \[ NH_4^+(aq) + OH^-(aq) + Li^+(aq) + Cl^-(aq) \rightarrow NH_3(g) + H_2O(\ell) + Li^+(aq) + Cl^-(aq) \]

5. Cancelling the spectator ions \( Li^+ \) and \( Cl^- \): The net ionic equation becomes: \[ NH_4^+(aq) + OH^-(aq) \rightarrow NH_3(g) + H_2O(\ell) \]

So the balanced net ionic equation is: \[ NH_4^+(aq) + OH^-(aq) \rightarrow NH_3(g) + H_2O(\ell) \]

Now, identifying the products that remain on the product side after spectator ions are canceled does indeed lead us to:

• \( NH_3(g) \) and \( H_2O(\ell) \)

Thus, the correct answer from the given options is: D. \( NH_3(g) + H_2O(\ell) \)