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AlekseyPX
3 years ago
9

What mass of precipitate forms when 185.5 ml of 0.533 m naoh is added to 627 ml of a solution that contains 15.8 g of aluminum s

ulfate per liter?
Chemistry
1 answer:
Law Incorporation [45]3 years ago
3 0

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 × \frac{1mol}{342,15g} = <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 × \frac{1molAl_{2}(SO_{4})_{3}}{6 moles 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 × \frac{2molAl(OH)_{3}}{6 moles 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>

<em></em>

I hope it helps!

<em> </em>

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