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22 May, 04:23

What is the limiting reactant when 7.0 grams of aluminum and 15.2 grams of oxygen react to produce aluminum oxide?

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  1. 22 May, 07:12
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    The number of moles of Al₂O₃ produced by Al are less it will limiting reactant.

    Explanation:

    Given dа ta:

    Mass of aluminium = 7.0 g

    Mass of oxygen = 15.2 g

    Limiting reactant = ?

    Solution:

    Chemical equation:

    4Al + 3O₂ → 2Al₂O₃

    Number of moles of Aluminium:

    Number of moles = mass / molar mass

    Number of moles = 7.0 g / 27 g/mol

    Number of moles = 0.3 mol

    Number of moles of oxygen:

    Number of moles = mass / molar mass

    Number of moles = 15.2 g / 32 g/mol

    Number of moles = 0.5 mol

    Now we will compare the moles of aluminium oxide with oxygen and Al.

    O₂ : Al₂O₃

    3 : 2

    0.5; 2/3*0.5 = 0.33 mol

    Al : Al₂O₃

    4 : 2

    0.3 : 2/4*0.3 = 0.15 mol

    The number of moles of Al₂O₃ produced by Al are less it will limiting reactant.
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