2 moles of pure `KClO_(3)` is decomposed to an extent of 66.6%. How many moles of `O_(2)` is released?

Answer

To solve the problem of how many moles of \( O_2 \) are released when 2 moles of pure \( KClO_3 \) are decomposed to an extent of 66.6%, we can follow these steps:

### Step 1: Write the decomposition reaction
The decomposition reaction of potassium chlorate (\( KClO_3 \)) can be represented as:
\[ 2 KClO_3 \rightarrow 2 KCl + 3 O_2 \]

### Step 2: Determine the moles of \( KClO_3 \) decomposed
Given that 2 moles of \( KClO_3 \) are decomposed to an extent of 66.6%, we can calculate the moles decomposed:
\[ \text{Moles decomposed} = \text{Total moles} \times \text{Extent of decomposition} \]
\[ \text{Moles decomposed} = 2 \text{ moles} \times 0.666 = 1.332 \text{ moles} \]

### Step 3: Use the stoichiometry of the reaction
From the balanced equation, we see that 2 moles of \( KClO_3 \) produce 3 moles of \( O_2 \). Therefore, we can set up a ratio to find out how many moles of \( O_2 \) are produced from the decomposed moles of \( KClO_3 \):
\[ \text{Moles of } O_2 = \left(\frac{3 \text{ moles } O_2}{2 \text{ moles } KClO_3}\right) \times \text{Moles decomposed} \]
\[ \text{Moles of } O_2 = \left(\frac{3}{2}\right) \times 1.332 = 1.998 \text{ moles} \]

### Step 4: Round to appropriate significant figures
Since we started with 2 moles of \( KClO_3 \), we can round our answer to 2 significant figures:
\[ \text{Moles of } O_2 \approx 2 \text{ moles} \]

### Final Answer
The number of moles of \( O_2 \) released is approximately **2 moles**.

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