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

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

Explanation:

Given data:

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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