Answer:
Here's what I get
Explanation:
1. In dilute NaOH
(a) Molecular equation
AlCl₃(aq) + 3NaOH(aq) ⟶ Al(OH)₃(s) + 3NaCl(aq)
(b) Ionic equation
You write the molecular formulas for solids, and you write the soluble ionic substances as ions.
Al³⁺(aq) + 3Cl⁻(aq) + 3Na⁺(aq) + 3OH⁻(aq) ⟶ Al(OH)₃(s) + 3Na⁺(aq) + 3Cl⁻(aq)
(c) Net ionic equation
To get the net ionic equation, you cancel the ions that appear on each side of the ionic equation.
Al³⁺(aq) + <u>3Cl⁻(aq)</u> + <u>3Na⁺(aq</u>) + 3OH⁻(aq) ⟶ Al(OH)₃(s) + <u>3Na⁺(aq)</u> + <u>3Cl⁻(aq)
</u>
The net ionic equation is
Al³⁺(aq) + 3OH⁻(aq) ⟶ Al(OH)₃(s)
2. In excess NaOH
The aluminium hydroxide reacts with excess hydroxide to form sodium tetrahydroxoaluminate(III).
(a) Molecular equation
AlCl₃(aq) + 4NaOH(aq) ⟶ NaAl(OH)₄(aq) + 3NaCl(aq)
(b) Ionic equation
Al³⁺(aq) + 3Cl⁻(aq) + 4Na⁺(aq) + 4OH⁻(aq) ⟶ Na⁺Al(OH)₄⁻(aq) + 3Na⁺(aq) + 3Cl⁻(aq)
(c) Net ionic equation
Al³⁺(aq) + 4OH⁻(aq) ⟶ Al(OH)₄⁻(aq)
