One mole of potassium chlorote is thermally decomposed and excess of aluminium is burnt in the gaseous product. How many mole(s) of aluminium oxide are formed?

Options

A. 1
B. 1.5
C. 2
D. 3

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Write the decomposition reaction of potassium chlorate (KClO3).
When potassium chlorate decomposes thermally, it produces potassium chloride (KCl) and oxygen gas (O2). The balanced reaction is:
\[ 2 \text{KClO}_3 \rightarrow 2 \text{KCl} + 3 \text{O}_2 \]

### Step 2: Determine the moles of oxygen produced from the decomposition of 1 mole of KClO3.
From the balanced equation, 2 moles of KClO3 produce 3 moles of O2. Therefore, for 1 mole of KClO3:
\[ \text{Moles of } O_2 = \frac{3 \text{ moles of } O_2}{2 \text{ moles of } KClO3} \times 1 \text{ mole of } KClO3 = 1.5 \text{ moles of } O_2 \]

### Step 3: Write the reaction of aluminum with oxygen to form aluminum oxide (Al2O3).
The reaction of aluminum with oxygen can be represented as:
\[ 4 \text{Al} + 3 \text{O}_2 \rightarrow 2 \text{Al}_2\text{O}_3 \]

### Step 4: Determine the moles of aluminum oxide produced from the available oxygen.
From the balanced equation, 3 moles of O2 produce 2 moles of Al2O3. Thus, we can set up a ratio:
\[ \text{Moles of } Al_2O_3 = \frac{2 \text{ moles of } Al_2O_3}{3 \text{ moles of } O_2} \times 1.5 \text{ moles of } O_2 \]

### Step 5: Calculate the moles of aluminum oxide produced.
Now, substituting the values:
\[ \text{Moles of } Al_2O_3 = 1.5 \times \frac{2}{3} = 1 \text{ mole of } Al_2O_3 \]

### Final Answer:
Thus, the number of moles of aluminum oxide (Al2O3) formed is **1 mole**.

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