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 find the percentage purity of KClO₃ based on the decomposition reaction that produces oxygen gas (O₂), we will follow these steps:

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

### Step 2: Determine the stoichiometric relationships
From the balanced equation, we can see that:
- 2 moles of KClO₃ produce 3 moles of O₂.
- Therefore, 1 mole of O₂ is produced by \(\frac{2}{3}\) moles of KClO₃.

### Step 3: Convert the volume of O₂ to moles
At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters (or 22400 mL).
To find the number of moles of O₂ produced from 'y' mL, we use the formula:
\[ \text{Moles of } O_2 = \frac{y \text{ mL}}{22400 \text{ mL/mole}} = \frac{y}{22400} \]

### Step 4: Calculate the moles of KClO₃
Using the stoichiometric relationship from Step 2, we can find the moles of KClO₃ that decomposed to produce 'y' mL of O₂:
\[ \text{Moles of KClO}_3 = \frac{2}{3} \times \text{Moles of } O_2 = \frac{2}{3} \times \frac{y}{22400} = \frac{2y}{67200} \]

### Step 5: Calculate the weight of KClO₃
Let the molar mass of KClO₃ be \(M\) g/mol. The weight of KClO₃ that decomposed can be calculated as:
\[ \text{Weight of KClO}_3 = \text{Moles of KClO}_3 \times M = \frac{2y}{67200} \times M \]

### Step 6: Calculate the percentage purity
The percentage purity of KClO₃ can be calculated using the formula:
\[ \text{Percentage Purity} = \left( \frac{\text{Weight of pure KClO}_3}{\text{Weight of sample}} \right) \times 100 \]
Substituting the values we have:
\[ \text{Percentage Purity} = \left( \frac{\frac{2y}{67200} \times M}{X} \right) \times 100 \]
This simplifies to:
\[ \text{Percentage Purity} = \frac{2yM}{67200X} \times 100 \]

### Final Expression
Thus, the final expression for the percentage purity of KClO₃ is:
\[ \text{Percentage Purity} = \frac{2yM \times 100}{67200X} \]