If half mole of oxygen combine with `Al` to form `Al_(2)O_(3)` the weight of `Al` used in the reaction is:

Options

A. `27 g`
B. `40.5 g`
C. `54 g`
D. `18 g`

Explanation

To solve the problem of how much aluminum (Al) is used when half a mole of oxygen (O₂) combines to form aluminum oxide (Al₂O₃), we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the formation of aluminum oxide from aluminum and oxygen is:
\[ 4 \, \text{Al} + 3 \, \text{O}_2 \rightarrow 2 \, \text{Al}_2\text{O}_3 \]

### Step 2: Determine the mole ratio
From the balanced equation, we can see that:
- 4 moles of Al react with 3 moles of O₂.

### Step 3: Calculate the moles of Al required for half a mole of O₂
Since we are given half a mole of O₂, we can use the mole ratio to find out how many moles of Al are needed:
- From the equation, 3 moles of O₂ require 4 moles of Al.
- Therefore, half a mole of O₂ will require:
\[ \text{Moles of Al} = \left(\frac{4 \, \text{moles Al}}{3 \, \text{moles O}_2}\right) \times 0.5 \, \text{moles O}_2 \]
\[ \text{Moles of Al} = \frac{4}{3} \times 0.5 = \frac{2}{3} \, \text{moles of Al} \]

### Step 4: Calculate the mass of Al used
Now, we need to convert the moles of Al to grams. The molar mass of aluminum (Al) is approximately 27 g/mol.
- Therefore, the mass of Al used is:
\[ \text{Mass of Al} = \text{Moles of Al} \times \text{Molar mass of Al} \]
\[ \text{Mass of Al} = \frac{2}{3} \, \text{moles} \times 27 \, \text{g/mol} \]
\[ \text{Mass of Al} = \frac{54}{3} = 18 \, \text{grams} \]

### Final Answer
The weight of aluminum used in the reaction is **18 grams**.

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