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.

what is the volume occupied by 40 grams of argon gas (ar) at standard conditions? what volume, in liters, is occupied by 1.5 x 1023 

law of conservation of matter