A sample of KCIO_3 on decomposition yielded 448 mL of oxygen gas at STP, then the weight of KCIO_3 originally taken was

Options

A. 0.815 g
B. 1.63 g
C. 3.27 g
D. 2.45 g

Answer

To solve the problem, we need to determine the weight of KClO3 that originally decomposed to yield 448 mL of oxygen gas at STP. Here’s a step-by-step solution:

### Step 1: Write the decomposition reaction of KClO3
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: Convert the volume of oxygen gas to liters
The volume of oxygen gas given is 448 mL. To convert this to liters:
\[ 448 \, \text{mL} = \frac{448}{1000} \, \text{L} = 0.448 \, \text{L} \]

### Step 3: Calculate the number of moles of oxygen gas
At STP (Standard Temperature and Pressure), 1 mole of gas occupies 22.4 L. To find the number of moles of oxygen gas produced:
\[ \text{Number of moles of } O_2 = \frac{\text{Volume of } O_2}{\text{Molar volume at STP}} = \frac{0.448 \, \text{L}}{22.4 \, \text{L/mol}} = 0.02 \, \text{moles} \]

### Step 4: Relate moles of oxygen to moles of KClO3
From the balanced equation, we see that 2 moles of KClO3 produce 3 moles of O2. Therefore, we can set up a ratio to find the moles of KClO3 needed to produce 0.02 moles of O2:
\[ \frac{2 \, \text{moles KClO}_3}{3 \, \text{moles } O_2} = \frac{x \, \text{moles KClO}_3}{0.02 \, \text{moles } O_2} \]
Cross-multiplying gives:
\[ x = \frac{2}{3} \times 0.02 = 0.01333 \, \text{moles KClO}_3 \]

### Step 5: Calculate the molar mass of KClO3
The molar mass of KClO3 is calculated as follows:
- K: 39.1 g/mol
- Cl: 35.5 g/mol
- O: 16.0 g/mol × 3 = 48.0 g/mol

Adding these together:
\[ \text{Molar mass of KClO}_3 = 39.1 + 35.5 + 48.0 = 122.6 \, \text{g/mol} \]

### Step 6: Calculate the mass of KClO3 required
Now we can find the mass of KClO3 required for 0.01333 moles:
\[ \text{Mass of KClO}_3 = \text{Number of moles} \times \text{Molar mass} = 0.01333 \, \text{moles} \times 122.6 \, \text{g/mol} = 1.63 \, \text{g} \]

### Final Answer
The weight of KClO3 originally taken was **1.63 grams**.

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