`27 g` of `Al` will react completely with…… `g` of `O_(2)`

Options

A. `8 g`
B. `10 g`
C. `24 g`
D. `49 g`

Explanation

To solve the problem of how much oxygen will react with 27 g of aluminum, we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction of aluminum with oxygen can be represented by the following 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 the number of moles of aluminum, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
The molar mass of aluminum (Al) is approximately 27 g/mol. Therefore, for 27 g of aluminum:
\[ \text{Number of moles of Al} = \frac{27 \text{ g}}{27 \text{ g/mol}} = 1 \text{ mole} \]

### Step 3: Determine the moles of oxygen required
From the balanced equation, we see that 4 moles of Al react with 3 moles of O2. We can set up a ratio to find out how many moles of O2 are needed for 1 mole of Al:
\[ \text{Moles of O}_2 = \left(\frac{3 \text{ moles O}_2}{4 \text{ moles Al}}\right) \times 1 \text{ mole Al} = \frac{3}{4} \text{ moles O}_2 \]

### Step 4: Calculate the mass of oxygen required
Now, we need to convert the moles of O2 into grams. The molar mass of O2 is approximately 32 g/mol. Therefore:
\[ \text{Mass of O}_2 = \text{Number of moles} \times \text{Molar mass} = \left(\frac{3}{4} \text{ moles}\right) \times 32 \text{ g/mol} \]
Calculating this gives:
\[ \text{Mass of O}_2 = \frac{3 \times 32}{4} = 24 \text{ g} \]

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

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