27g of Al was treated with NaOH solution when a white gelatinous precipitate was obtained which upon strong heating gave an oxide. The amount of oxide (in g ) is

Explanation

To solve the problem step by step, let's break down the information given and perform the necessary calculations.

### Step 1: Determine the moles of Aluminum (Al)
We start with 27 g of Aluminum (Al). The molar mass of Aluminum is approximately 27 g/mol.

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

### Step 2: Write the reaction with NaOH
When Aluminum reacts with NaOH, it forms a white gelatinous precipitate of Aluminum hydroxide (Al(OH)₃):

\[ \text{2 Al} + \text{6 NaOH} + \text{6 H₂O} \rightarrow \text{2 Al(OH)₃} + \text{6 Na} \]

From the reaction, we can see that 2 moles of Aluminum produce 2 moles of Al(OH)₃.

### Step 3: Determine the amount of Aluminum hydroxide produced
Since we have 1 mole of Aluminum, we will produce 1 mole of Aluminum hydroxide:

\[ \text{Moles of Al(OH)₃} = 1 \text{ mol} \]

### Step 4: Calculate the mass of Aluminum hydroxide (Al(OH)₃)
The molar mass of Al(OH)₃ can be calculated as follows:

\[ \text{Molar mass of Al(OH)₃} = \text{(1 × 27)} + \text{(3 × 16)} + \text{(3 × 1)} = 27 + 48 + 3 = 78 \text{ g/mol} \]

Now, we can calculate the mass of Al(OH)₃ produced:

\[ \text{Mass of Al(OH)₃} = \text{moles} \times \text{molar mass} = 1 \text{ mol} \times 78 \text{ g/mol} = 78 \text{ g} \]

### Step 5: Heating Aluminum hydroxide to form Aluminum oxide (Al₂O₃)
When Aluminum hydroxide is heated, it decomposes to form Aluminum oxide (Al₂O₃):

\[ \text{2 Al(OH)₃} \rightarrow \text{Al₂O₃} + \text{3 H₂O} \]

From the reaction, we see that 2 moles of Al(OH)₃ yield 1 mole of Al₂O₃. Since we have 1 mole of Al(OH)₃, we will produce:

\[ \text{Moles of Al₂O₃} = \frac{1 \text{ mol Al(OH)₃}}{2} = 0.5 \text{ mol} \]

### Step 6: Calculate the mass of Aluminum oxide (Al₂O₃)
The molar mass of Al₂O₃ is calculated as follows:

\[ \text{Molar mass of Al₂O₃} = \text{(2 × 27)} + \text{(3 × 16)} = 54 + 48 = 102 \text{ g/mol} \]

Now, we can calculate the mass of Al₂O₃ produced:

\[ \text{Mass of Al₂O₃} = \text{moles} \times \text{molar mass} = 0.5 \text{ mol} \times 102 \text{ g/mol} = 51 \text{ g} \]

### Final Answer
The amount of Aluminum oxide (Al₂O₃) produced is **51 g**.

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