One mole of `KClO_(3)` is heated in presence of `MnO_(2)`. The produced oxygen is used in burning of Al. Then oxide of Al that will be formed.

Options

A. 2 mole
B. 1 mole
C. 4 mole
D. 3 mole

Answer

To solve the problem step by step, we will follow the chemical reactions involved and the stoichiometry of the reactants and products.

### Step 1: Decomposition of KClO₃
When potassium chlorate (KClO₃) is heated in the presence of manganese dioxide (MnO₂), it decomposes to form potassium chloride (KCl) and oxygen (O₂). The balanced equation for this reaction is:

\[ 2 \text{KClO}_3 \xrightarrow{\text{heat}, \text{MnO}_2} 2 \text{KCl} + 3 \text{O}_2 \]

From the equation, we can see that 2 moles of KClO₃ produce 3 moles of O₂. Therefore, 1 mole of KClO₃ will produce:

\[ \text{O}_2 \text{ produced} = \frac{3}{2} \text{ moles of O}_2 \]

### Step 2: Reaction of Oxygen with Aluminum
Now, the produced oxygen (O₂) will react with aluminum (Al) to form aluminum oxide (Al₂O₃). The balanced equation for this reaction is:

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

### Step 3: Determine Moles of Aluminum Oxide Formed
From the reaction, we can see that 3 moles of O₂ react with 4 moles of Al to produce 2 moles of Al₂O₃. We need to find out how many moles of Al₂O₃ can be produced from the oxygen generated from 1 mole of KClO₃.

Since 1 mole of KClO₃ produces \( \frac{3}{2} \) moles of O₂, we can set up a proportion to find the moles of Al₂O₃ produced:

Using the stoichiometry from the reaction of Al with O₂:

\[ \frac{3 \text{ moles O}_2}{2 \text{ moles Al}_2\text{O}_3} \]

We can calculate the moles of Al₂O₃ produced from \( \frac{3}{2} \) moles of O₂:

\[ \text{Moles of Al}_2\text{O}_3 = \frac{2}{3} \times \frac{3}{2} = 1 \text{ mole of Al}_2\text{O}_3 \]

### Conclusion
Thus, the oxide of aluminum that will be formed is:

\[ \text{Al}_2\text{O}_3 \]