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 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.