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28 July, 20:41

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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  1. 28 July, 20:45
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    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.
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