Answer to Question #120903 in Organic Chemistry for Uche

Question

(A) Aluminum chloride reacts with sodium hydroxide to produce an aluminum hydroxide precipitate. Write a balance chemical reaction.
(B) How many moles aluminum hydroxide and sodium chloride would be produced if 2.0 molesof sodium hydroxide are reacted with excess aluminum chloride?

Answer 1

Solution:

Part A:

Aluminum chloride react with sodium hydroxide to produce aluminium hydroxide and sodium chloride.

The balanced chemical equation:

AlCl₃(aq) + 3NaOH(aq) → Al(OH)₃(s) + 3NaCl(aq)

Ionic equation: Al³⁺(aq) + 3Cl^-(aq) + 3Na⁺(aq) + 3OH^-(aq) → Al(OH)₃(s) + 3Na⁺(aq) + 3Cl^-(aq)

Net ionic equation: Al³⁺(aq) + 3OH^-(aq) → Al(OH)₃(s)

Part B:

The balanced chemical equation:

AlCl₃(aq) + 3NaOH(aq) → Al(OH)₃(s) + 3NaCl(aq)

According to the chemical equation:

1) n(NaOH) = n(NaCl) = 2.0 mol

Moles of NaCl = 2.0 mol

2) n(NaOH)/3 = n(Al(OH)₃)

Moles of Al(OH)3 = Moles of NaOH / 3 = (2.0 mol) / 3 = 0.67 mol

0.67 mol of aluminum hydroxide (Al(OH)₃) and 2.0 mol of sodium chloride (NaCl) would be produced.