If `1 1/2` moles of oxygen combine with `Al` to form `Al_(2)O_(3)` the weight of `Al` used in the reaction is `(Al=27)`

Options

A. 27 g
B. 54 g
C. 49.5 g
D. 31 g

Explanation

To solve the problem of how much aluminum (Al) is used when 1.5 moles of oxygen (O2) combine to form aluminum oxide (Al2O3), we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction of aluminum with oxygen to form aluminum oxide 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 needed for 1.5 moles of O2
We need to find out how many moles of Al are required for 1.5 moles of O2. Using the mole ratio from the balanced equation:
\[ \text{Moles of Al} = \left( \frac{4 \text{ moles of Al}}{3 \text{ moles of O}_2} \right) \times 1.5 \text{ moles of O}_2 \]
Calculating this gives:
\[ \text{Moles of Al} = \frac{4}{3} \times 1.5 = 2 \text{ moles of Al} \]

### Step 4: Calculate the mass of Al used
Now, we need to calculate the mass of aluminum used in the reaction. The molar mass of aluminum (Al) is given as 27 g/mol. Therefore, the mass of Al can be calculated using the formula:
\[ \text{Mass} = \text{Moles} \times \text{Molar Mass} \]
Substituting the values:
\[ \text{Mass of Al} = 2 \text{ moles} \times 27 \text{ g/mol} = 54 \text{ g} \]

### Final Answer
The weight of aluminum used in the reaction is **54 grams**.

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