18.625g KCl is formed due to decomposition of `KClO_(4)` in reaction
`KClO_(4)(s)rarr KCl(s)+2O_(2)(g)`
Find volume of `O_(2)` obtained at STP[Atomic mass K = 39 Cl = 35.5]

Answer

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

### Step 1: Write the decomposition reaction
The decomposition of potassium perchlorate (KClO₄) is given by the reaction:
\[ \text{KClO}_4 (s) \rightarrow \text{KCl} (s) + 2 \text{O}_2 (g) \]

### Step 2: Calculate the molar mass of KCl
The molar mass of KCl can be calculated using the atomic masses provided:
- Atomic mass of K = 39 g/mol
- Atomic mass of Cl = 35.5 g/mol

Thus, the molar mass of KCl is:
\[ \text{Molar mass of KCl} = 39 + 35.5 = 74.5 \, \text{g/mol} \]

### Step 3: Calculate the number of moles of KCl formed
We are given that 18.625 g of KCl is formed. To find the number of moles of KCl, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]

Substituting the values:
\[ \text{Number of moles of KCl} = \frac{18.625 \, \text{g}}{74.5 \, \text{g/mol}} \approx 0.25 \, \text{mol} \]

### Step 4: Determine the moles of O₂ produced
From the balanced equation, we see that 1 mole of KCl produces 2 moles of O₂. Therefore, if we have 0.25 moles of KCl, the moles of O₂ produced will be:
\[ \text{Moles of O}_2 = 0.25 \, \text{mol KCl} \times 2 \, \frac{\text{mol O}_2}{\text{mol KCl}} = 0.5 \, \text{mol O}_2 \]

### Step 5: Calculate the volume of O₂ at STP
At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters. Therefore, the volume of 0.5 moles of O₂ can be calculated as:
\[ \text{Volume of O}_2 = 0.5 \, \text{mol} \times 22.4 \, \text{L/mol} = 11.2 \, \text{L} \]

### Final Answer
The volume of O₂ obtained at STP is **11.2 liters**.

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