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

54 g of Aluminium were oxidized

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

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