27g of Al will react completely with

Options

A. 24 g of `O_(2)`
B. 0.75 mole of `O_(2)`
C. 16.8 L of `O_(2)` at STP
D. 1 mole of `O_(2)`

Answer

To solve the question of how much oxygen reacts completely with 27 g of aluminum, we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between aluminum (Al) and oxygen (O₂) to form aluminum oxide (Al₂O₃) can be represented by the balanced equation:

\[ 4 \text{Al} + 3 \text{O}_2 \rightarrow 2 \text{Al}_2\text{O}_3 \]

### Step 2: Calculate the number of moles of aluminum
To find out how many moles of aluminum we have in 27 g, we use the formula:

\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]

The molar mass of aluminum (Al) is approximately 27 g/mol. Therefore:

\[ \text{Number of moles of Al} = \frac{27 \text{ g}}{27 \text{ g/mol}} = 1 \text{ mole} \]

### Step 3: Use the stoichiometry of the balanced equation
From the balanced equation, we see that 4 moles of aluminum react with 3 moles of oxygen. Therefore, the ratio of aluminum to oxygen is:

\[ \frac{4 \text{ moles Al}}{3 \text{ moles O}_2} \]

To find out how many moles of oxygen are needed for 1 mole of aluminum, we set up a proportion:

\[ \text{Moles of O}_2 = \frac{3 \text{ moles O}_2}{4 \text{ moles Al}} \times 1 \text{ mole Al} = 0.75 \text{ moles O}_2 \]

### Step 4: Calculate the mass of oxygen required
Now we need to calculate the mass of oxygen required. The molar mass of O₂ is approximately 32 g/mol (16 g/mol for each oxygen atom). Therefore:

\[ \text{Mass of O}_2 = \text{Number of moles} \times \text{Molar mass} = 0.75 \text{ moles} \times 32 \text{ g/mol} = 24 \text{ g} \]

### Conclusion
Thus, 27 g of aluminum will react completely with **24 g of oxygen**.

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