Question

(LO 3N, 4G, 4O) Aluminum sulfate is also involved in dying fabrics. The gelatinous precipitate formed in the reaction with dilute sodium hydroxide helps dye adhere to fabric: Al₂(SO₄)₃(aq) + 6 NaOH(aq) → 2 Al(OH)₃(?) + 3 Na₂SO₄(?) (Balanced) If 5.6x10⁻³ mol of Al₂(SO₄)₃ is added to 25.0 mL of 0.088 M NaOH, how much gelatinous precipitate is formed?

Answer 1

9.4 × 10⁻³ mg (0.73 mmoles) of Al(OH)₃ is formed

Explanation:

We have the following chemical reaction:

Al₂(SO₄)₃(aq) + 6 NaOH(aq) → 2 Al(OH)₃(s) + 3 Na₂SO₄(aq)

The precipitate mentioned by the problem is aluminium hydroxide Al(OH)₃.

Now to determine the number of moles of sodium hydroxide NaOH we use the following formula:

molar concentration =  number of moles / volume

number of moles = molar concentration × volume

number of moles of NaOH = 0.088 M × 25 mL = 2.2 mmoles

number of moles of Al₂(SO₄)₃ = 5.6 × 10⁻³ moles = 5.6 mmoles (found in the  problem text)

We see from the chemical reaction that 1 mole of Al₂(SO₄)₃ requires 6 moles of NaOH so 5.6 mmoles of Al₂(SO₄)₃ would require 6 times more NaOH which is 33.6 mmoles and we have only 2.2 mmoles. The limiting reactant will be NaOH.

Now we devise the following reasoning:

if        6 mmoles of NaOH produces 2 mmoles of Al(OH)₃

then  2.2 mmoles of NaOH produces X mmoles of Al(OH)₃

X = (2.2 × 2) / 6 = 0.73 mmoles of Al(OH)₃

mass of Al(OH)₃ = number of moles / molecular weight

mass of Al(OH)₃ = 0.73 / 78

mass of Al(OH)₃ =  9.4 × 10⁻³ mg

Learn more:

precipitation reaction

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