245 g impure sample of `KClO_3` on heating gives `12gO_2(g)` according to `2KClO_3(s)rarr2KCl(s)+3 O_2(g)` Calculate % purity of sample ?

Answer

To calculate the percentage purity of the impure sample of KClO3, we can follow these steps:

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

### Step 2: Calculate the moles of O2 produced
We know that the mass of O2 produced is 12 g. To find the number of moles of O2, we use the molar mass of O2, which is 32 g/mol.
\[ \text{Moles of O}_2 = \frac{\text{mass of O}_2}{\text{molar mass of O}_2} = \frac{12 \, \text{g}}{32 \, \text{g/mol}} = 0.375 \, \text{mol} \]

### Step 3: Relate moles of O2 to moles of KClO3
From the balanced equation, we see that 3 moles of O2 are produced from 2 moles of KClO3. Therefore, we can find the moles of KClO3 that produced 0.375 moles of O2:
\[ \text{Moles of KClO}_3 = \frac{2}{3} \times \text{moles of O}_2 = \frac{2}{3} \times 0.375 = 0.25 \, \text{mol} \]

### Step 4: Calculate the mass of KClO3
Now, we can calculate the mass of KClO3 using its molar mass. The molar mass of KClO3 is approximately 122.5 g/mol.
\[ \text{Mass of KClO}_3 = \text{moles of KClO}_3 \times \text{molar mass of KClO}_3 = 0.25 \, \text{mol} \times 122.5 \, \text{g/mol} = 30.625 \, \text{g} \]

### Step 5: Calculate the percentage purity of the sample
The percentage purity of the sample can be calculated using the formula:
\[ \text{Percentage purity} = \left( \frac{\text{mass of pure KClO}_3}{\text{total mass of sample}} \right) \times 100 \]
Substituting the values we have:
\[ \text{Percentage purity} = \left( \frac{30.625 \, \text{g}}{245 \, \text{g}} \right) \times 100 \approx 12.5\% \]

### Final Answer
The percentage purity of the sample is **12.5%**.

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