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20 April, 19:02

Aluminum oxide forms when aluminum reacts with oxygen. 4Al (s) + 3O2 (g) → 2Al2O3 (s) A mixture of 82.49 g of aluminum (M. W. = 26.98 g/mol) and 117.65 g of oxygen (M. W. = 32.00 g/mol) is allowed to react. What mass of aluminum oxide (M. W. = 101.96 g/mol) can be formed?

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  1. 20 April, 19:19
    0
    155.4 g of Al₂O₃

    Explanation:

    The reaction is:

    4Al (s) + 3O₂ (g) → 2Al₂O₃ (s)

    To determine the mass of aluminum oxide that is formed we need to know the limiting reagent. Let's calculate the moles of each by the molar mass

    Mass / molar mass = Moles

    82.49 g / 26.98 g/mol = 3.05 moles of Al

    117.65 g / 32 g/mol = 3.67 moles of oxygen

    Let's try the oxygen. Ratio is 3:4.

    3 moles of O₂ need 4 moles of Al to react

    Therefore 3.67 moles of O₂ will react with (3.67. 4) / 3 = 4.90 moles

    We only have 3.05 moles of Al, so the Al is the limiting reactant

    Now, we work with stoichiometry

    4 moles of Al can produce 2 moles of Al₂O₃

    3.05 moles of Al will produce (3.05.2) / 4 = 1.52 moles of Al₂O₃

    We convert the moles to mass: 1.52 mol. 101.96 g / 1mol = 155.4 g
  2. 20 April, 19:24
    0
    155.9 grams of aluminium oxide can be formed

    Explanation:

    Step 1: Data given

    Mass of aluminium = 82.49 grams

    Molar mass of aluminium = 26.98 g/mol

    Mass of oxygen = 117.65 grams

    Molar mass of oxygen = 32.0 g/mol

    Molar mass of aluminium oxide = 101.96 g/mol

    Step 2: The balanced equation

    4Al + 3O2 → 2Al2O3

    Step 3: Calculate moles

    Moles = mass / molar mass

    Moles aluminium = 82.49 grams / 26.98 g/mol

    Moles aluminium = 3.057 moles

    Moles oxygen = 117.65 grams / 32.0 g/mol

    Moles oxygen = 3.677 moles

    Step 4: Calculate limiting reactant

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

    Aluminium is the limiting reactant. It will completely be consumed (3.057 moles). Oxygen is in excess. There will react 3/4 * 3.057 = 2.293 moles

    There will remain 3.677 - 2.293 = 1.384 moles

    Step 5: Calculate moles aluminium oxide

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

    For 3.057 moles Al we'll have 3.057/2 = 1.529 moles

    Step 6: Calculate mass aluminium oxide

    Mass aluminium oxide = 1.529 moles * 101.96 g/mol

    Mass aluminium oxide = 155.9 grams

    155.9 grams of aluminium oxide can be formed
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