If 355 mL of 1.50 M aluminum nitrate is added to an excess of sodium sulfate, how many grams of aluminum sulfate will be produced?

Question

Chemistry

Paityn Mueller

10 December, 13:38

If 355 mL of 1.50 M aluminum nitrate is added to an excess of sodium sulfate, how many grams of aluminum sulfate will be produced?

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Answers (1)

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Stella Lester · Answer 1

Answer is: 91.1 grams of aluminum sulfate.

Balanced chemical reaction: 2Al (NO₃) ₃ + 3Na₂SO₄→ Al₂ (SO₄) ₃ + 6NaNO₃.

V (Al (NO₃) ₃) = 355 mL : 1000 mL/L = 0.355 L.

c (Al (NO₃) ₃) = 1.5 mol/L.

n (Al (NO₃) ₃) = V (Al (NO₃) ₃) · c (Al (NO₃) ₃).

n (Al (NO₃) ₃) = 0,355 L · 1.5 mol/L.

n (Al (NO₃) ₃) = 0.5325 mol.

From chemical reaction: n (Al (NO₃) ₃) : n (Al₂ (SO₄) ₃) = 2 : 1.

n (Al₂ (SO₄) ₃) = 0.266 mol.

m (Al₂ (SO₄) ₃) = 0.266 mol · 342.15 g/mol.

m (Al₂ (SO₄) ₃) = 91.1 g.