Answer:
2,57 g of precipitate.
Explanation:
For the reaction:
6 NaOH + Al₂(SO₄)₃ → 2 Al(OH)₃ + 3 Na₂SO₄
The precipitate is Al(OH)₃.
185,5mL of 0,533M NaOH are:
0,1855L × 0,533M = <em>0,0989 moles NaOH</em>
Moles of Al₂(SO₄)₃ are:
15,8g × = <em>0,0462 moles Al₂(SO₄)₃</em>
For the total reaction of 0,0989 moles NaOH with Al₂(SO₄)₃ you need:
0,0989moles NaOH × = <em>0,0165 moles Al₂(SO₄)₃</em>
As you have <em>0,0462 moles Al₂(SO₄)₃ </em>the limiting reactant is NaOH.
0,0989 moles of NaOH produce:
0,0989moles NaOH × = <em>0,0330 moles of Al(OH)₃</em>
These moles are:
0,0330 moles of Al(OH)₃ × (78 g/mol) = <em>2,57 g of Al(OH)₃ ≡ mass of precipitate</em>
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I hope it helps!
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