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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