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

Options

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

Explanation

To solve the problem of how much oxygen (O₂) will react with 27 g of aluminum (Al), 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 out how many moles of aluminum are present in 27 g, we use the formula:

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

The molecular 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 stoichiometry to find moles of O₂ needed
From the balanced equation, we see that 4 moles of aluminum react with 3 moles of oxygen. Thus, we can set up a proportion 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: Convert moles of O₂ to grams
Now we need to convert the moles of O₂ into grams. The molecular mass of O₂ is approximately 32 g/mol. Therefore, we calculate the mass of O₂ required:

\[ \text{Mass of O}_2 = \text{Number of moles} \times \text{Molecular mass} \]

\[ \text{Mass of O}_2 = \frac{3}{4} \text{ moles} \times 32 \text{ g/mol} = 24 \text{ g} \]

### Final Answer
Thus, 27 g of aluminum will react completely with **24 g of O₂**.

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