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24 October, 23:23

Aluminum and oxygen react according to the following equation: 4Al (s) 3O2 (g) - -> 2Al2O3 (s) What mass of Al2O3, in grams, can be made by reacting 4.6 g Al with excess oxygen

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  1. 24 October, 23:27
    0
    8.66 g of Al₂O₃ will be produced

    Explanation:

    4Al (s) + 3O₂ (g) → 2Al₂O₃ (s)

    This is the reaction.

    Problem statement says, that the O₂ is in excess, so the limiting reactant is the Al. Let's determine the moles we used.

    4.6 g / 26.98 g/mol = 0.170 moles

    Ratio is 4:2.

    4 moles of aluminum can produce 2 moles of Al₂O₃

    0.170 moles of Al, may produce (0.170.2) / 4 = 0.085 moles

    Let's convert the moles of Al₂O₃ to mass.

    0.085 mol. 101.96 g/mol = 8.66 g
  2. 24 October, 23:45
    0
    There will be made 8.7 grams of Al2O3

    Explanation:

    Step 1: Data given

    Mass of Al = 4.6 grams

    Molar mass Al = 26.98 g/mol

    Oxygen is in excess

    Molar mass of Al2O3 = 101.96 g/mol

    Step 2: The balanced equation

    4Al (s) + 3O2 (g) → 2Al2O3 (s)

    Step 3: Calculate moles Al

    Moles Al = Mass Al / molar mass Al

    Moles Al = 4.6 grams / 26.98 g/mol

    Moles Al = 0.1705 moles

    Step 4: Calculate moles Al2O3

    For 4 moles Al we need 3 moles O2 to produce 2 moles of Al2O3

    For 0.1705 moles Al we'll have 0.1705 / 2 = 0.08525 moles Al2O3

    Step 5: Calculate mass Al2O3

    Mass Al2O3 = moles Al2O3 * molar mass Al2O3

    Mass Al2O3 = 0.08525 moles * 101.96 g/mol

    Mass Al2O3 = 8.7 grams

    There will be made 8.7 grams of Al2O3
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