How much potassium chlorate should be heated to produce 2.24L of oxygen at NTP?

Answer

To determine how much potassium chlorate (KClO3) should be heated to produce 2.24 L of oxygen (O2) at Normal Temperature and Pressure (NTP), we can follow these steps:

### Step 1: Write the balanced chemical equation for the decomposition of potassium chlorate.
The decomposition reaction of potassium chlorate can be represented as follows:

\[ 2 \text{KClO}_3 \rightarrow 2 \text{KCl} + 3 \text{O}_2 \]

This equation tells us that 2 moles of potassium chlorate produce 3 moles of oxygen gas.

### Step 2: Determine the volume of oxygen produced from the decomposition of potassium chlorate.
At NTP, 1 mole of any gas occupies 22.4 liters. Therefore, 3 moles of oxygen will occupy:

\[ 3 \text{ moles} \times 22.4 \text{ L/mole} = 67.2 \text{ L} \]

### Step 3: Calculate the amount of potassium chlorate needed to produce 2.24 L of oxygen.
We need to find out how many grams of potassium chlorate are required to produce 2.24 L of oxygen. We can set up a proportion based on the volume of oxygen produced:

\[ \frac{245 \text{ g KClO}_3}{67.2 \text{ L O}_2} = \frac{x \text{ g KClO}_3}{2.24 \text{ L O}_2} \]

Where \(x\) is the mass of potassium chlorate we want to find.

### Step 4: Solve for \(x\).
Cross-multiplying gives:

\[ x = \frac{245 \text{ g KClO}_3 \times 2.24 \text{ L O}_2}{67.2 \text{ L O}_2} \]

Calculating this:

\[ x = \frac{245 \times 2.24}{67.2} \]

\[ x = \frac{549.8}{67.2} \approx 8.71 \text{ g KClO}_3 \]

### Final Answer:
To produce 2.24 L of oxygen at NTP, approximately **8.71 grams of potassium chlorate** should be heated.

---