The weight of oxygen required to completely react with with 27 gms of 'Al' is

Options

A. 8 gm
B. 16 gm
C. 32 gm
D. 24 gm

Explanation

To find the weight of oxygen required to completely react with 27 grams of aluminum (Al), we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between aluminum and 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: Determine the molar masses
- The molar mass of aluminum (Al) is 27 g/mol.
- The molar mass of oxygen (O2) is 32 g/mol (16 g/mol for each oxygen atom).

### Step 3: Calculate the moles of aluminum
To find the moles of aluminum in 27 grams, we use the formula:

\[ \text{Moles of Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}} = \frac{27 \text{ g}}{27 \text{ g/mol}} = 1 \text{ mol} \]

### Step 4: Use the stoichiometry of the reaction
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:

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

### Step 5: Calculate the moles of oxygen needed
Using the moles of aluminum we calculated:

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

### Step 6: Convert moles of oxygen to grams
Now, we convert the moles of oxygen back to grams using its molar mass:

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

### Conclusion
The weight of oxygen required to completely react with 27 grams of aluminum is **24 grams**.

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