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25 December, 14:41

How many grams of aluminum sulfate form when 3.90 grams of aluminum is placed in 13.65 grams of sulfuric acid?

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  1. 25 December, 17:58
    0
    Mass = 17.12 g

    Explanation:

    Given dа ta:

    Mass of Al = 3.90 g

    Mass of H₂SO₄ = 13.65

    Mass of aluminium sulfate = ?

    Solution:

    Chemical equation:

    3H₂SO₄ + 2Al → Al₂ (SO₄) ₃ + 3H₂

    Now we will calculate the number of moles of each reactant.

    Moles of H₂SO₄:

    Number of moles = mass / molar mass

    Molar mass of H₂SO₄ = 98.079 g/mol

    Number of moles = 13.65 g / 98.079 g/mol

    Number of moles = 0.14 mol

    Moles of Al:

    Number of moles = mass / molar mass

    molar mass of Al = 27 g/mol

    Number of moles = 3.90 g / 27 g/mol

    Number of moles = 0.14 mol

    Now we will compare the moles of aluminium sulfate with sulfuric acid and aluminium.

    H₂SO₄ : Al₂ (SO₄) ₃

    3 : 1

    0.14 : 1/3*0.14 = 0.05

    Al : Al₂ (SO₄) ₃

    2 : 1

    0.14 : 1/2*0.14 = 0.07

    The number of moles of aluminium sulfate produced by sulfuric acid are less so it will limiting reactant and limit the amount of aluminium sulfate.

    Mass of aluminium sulfate:

    Mass = number of moles * molar mass

    molar mass of aluminium sulfate = 342.15 g/mol

    Mass = 0.05 mol * 342.15 g/mol

    Mass = 17.12 g
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