x' g of `KCIO_3` on decomposition gives 'y' ml of `O_2` at STP. The percentage purity of `KCIO_3` would be :

Options

A. `(yxx 2 xx M)/(22.4 xx 3 xx x)` x100
B. `(y xx 3 xx M)/(224 xx 2 xx x)` x100
C. `( y xx 2 xx M )/(22400xx 3 xx x)` x100
D. `( y xx 3 xx M)/(22400 xx 2 xx x)` x100

Answer

To solve the problem of finding the percentage purity of KClO3 based on the volume of oxygen produced during its decomposition, we can follow these steps:

### Step-by-Step Solution:

1. **Write the Decomposition Reaction**:
The decomposition of potassium chlorate (KClO3) can be represented by the following balanced chemical equation:
\[ 2 \text{KClO}_3 \rightarrow 2 \text{KCl} + 3 \text{O}_2 \]

2. **Determine Moles of Oxygen Produced**:
At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters or 22400 ml. Given that 'y' ml of O2 is produced, the number of moles of O2 can be calculated as:
\[ \text{Moles of } O_2 = \frac{y \text{ ml}}{22400 \text{ ml/mol}} = \frac{y}{22400} \text{ moles} \]

3. **Relate Moles of KClO3 to Moles of O2**:
From the balanced equation, we see that 2 moles of KClO3 produce 3 moles of O2. Therefore, the moles of KClO3 that would produce the calculated moles of O2 can be found using the ratio:
\[ \text{Moles of KClO}_3 = \frac{2}{3} \times \text{Moles of } O_2 = \frac{2}{3} \times \frac{y}{22400} = \frac{2y}{67200} \text{ moles} \]

4. **Calculate the Moles of KClO3 from Given Mass**:
The number of moles of KClO3 based on the mass 'x' grams is given by:
\[ \text{Moles of KClO}_3 = \frac{x \text{ grams}}{122.5 \text{ g/mol}} = \frac{x}{122.5} \text{ moles} \]

5. **Set Up the Purity Equation**:
The percentage purity of KClO3 can be calculated using the formula:
\[ \text{Percentage Purity} = \left( \frac{\text{Moles of KClO}_3 \text{ that decomposed}}{\text{Total Moles of KClO}_3} \right) \times 100 \]
Substituting the values we found:
\[ \text{Percentage Purity} = \left( \frac{\frac{2y}{67200}}{\frac{x}{122.5}} \right) \times 100 \]

6. **Simplify the Expression**:
Rearranging the equation gives:
\[ \text{Percentage Purity} = \left( \frac{2y \times 122.5}{67200 \times x} \right) \times 100 \]
This simplifies to:
\[ \text{Percentage Purity} = \frac{2y \times 122.5 \times 100}{67200 \times x} \]

### Final Result:
The final formula for the percentage purity of KClO3 is:
\[ \text{Percentage Purity} = \frac{2y \times 122.5 \times 100}{67200 \times x} \]