**Calculation of the Weight of Al Used**

To determine the weight of aluminum (Al) used in the reaction, we need to consider the balanced chemical equation and the molar ratios between oxygen (O2) and aluminum oxide (Al2O3).

The balanced chemical equation for the reaction between aluminum and oxygen to form aluminum oxide is:

4 Al + 3 O2 → 2 Al2O3

From the balanced equation, we can deduce the following molar ratios:

- 4 moles of Al react with 3 moles of O2
- 2 moles of Al2O3 are formed from 4 moles of Al

Since we are given that half a mole of O2 combines with Al to form Al2O3, we can use these molar ratios to calculate the weight of Al used.

**Molar Mass of O2 and Al2O3**

Before proceeding with the calculation, we need to determine the molar mass of O2 and Al2O3.

- The molar mass of O2 is 32 g/mol (16 g/mol for each oxygen atom, multiplied by 2).
- The molar mass of Al2O3 is 101.96 g/mol (2 × 26.98 g/mol for aluminum + 3 × 16 g/mol for oxygen).

**Calculating the Moles of O2 and Al2O3**

Since we are given that half a mole of O2 combines with Al, we can calculate the moles of O2 and Al2O3 involved in the reaction.

- Moles of O2 = 0.5 mol
- Moles of Al2O3 = (0.5 mol O2) / (3 mol O2/2 mol Al2O3) = 0.1667 mol

**Calculating the Moles of Al**

Using the molar ratio between Al and Al2O3, we can calculate the moles of Al required.

- Moles of Al = (0.1667 mol Al2O3) × (4 mol Al/2 mol Al2O3) = 0.3334 mol

**Calculating the Weight of Al Used**

Finally, we can determine the weight of Al used by multiplying the moles of Al by its molar mass.

- Weight of Al = (0.3334 mol Al) × (26.98 g/mol Al) = 8.99 g

Therefore, the weight of aluminum used in the reaction is approximately 8.99 grams.