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zlopas [31]
3 years ago
7

AlCl3 and NaOH Balanced Equation Total Ionic Equation Net Ionic Equation

Chemistry
1 answer:
Sindrei [870]3 years ago
7 0

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)  

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