Question

When aluminum oxidizes in air, it forms aluminum oxide (Al2O3):

4Al(s)+3O2(g)→2Al2O3(s)

A 102 g sheet of aluminum oxide formed completely in excess oxygen.

How many grams of aluminum were oxidized?


27 g


51 g


54 g


102 g


54 is correct answer

Answer 1

54 g of Aluminium were oxidized

Step-by-step explanation:

The equation for the reaction:

4Al(s)+3O₂(g)→2Al₂O₃(s)

We are required to calculate the mass of Aluminium that reacted;

Step 1: Calculating the number of moles Aluminum oxide

Mass of Aluminium oxide = 102 g

Molar mass of Al₂O₃ = 101.96 g/mol

Number of moles = Mass ÷ Molar mass

= 102 g ÷ 101.96 g/mol

= 1 mole

Step 2: Number of moles of Aluminium oxidized

From the equation, the mole ratio of Al to Al₂O₃ is 4 : 2

Therefore; Moles of Al is (4/2)(1) is 2 moles

2 moles of Aluminium were oxidized

Step 3: Mass of Aluminium oxidized

Mass = Number of moles × Molar mass

Molar mass of Aluminium is 27 g/mol

Therefore;

Mass of Aluminium = 2 moles × 27 g/mol

= 54 g

Therefore, 54 g of Aluminium were oxidized