What volume of oxygen at `18^(@)C and 750 mm` pressure can be obtained from 10 g of potassium chlorate?

Explanation

To find the volume of oxygen gas that can be obtained from the decomposition of 10 g of potassium chlorate (KClO3) at 18°C and 750 mmHg, we will follow these steps:

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

### Step 2: Calculate the molar mass of potassium chlorate (KClO3).
- Potassium (K) = 39 g/mol
- Chlorine (Cl) = 35.5 g/mol
- Oxygen (O) = 16 g/mol

Calculating the molar mass:
\[ \text{Molar mass of KClO}_3 = 39 + 35.5 + (3 \times 16) = 39 + 35.5 + 48 = 122.5 \text{ g/mol} \]

### Step 3: Calculate the number of moles of KClO3 in 10 g.
Using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
\[ \text{Number of moles of KClO}_3 = \frac{10 \text{ g}}{122.5 \text{ g/mol}} \approx 0.0816 \text{ mol} \]

### Step 4: Use the stoichiometry of the reaction to find moles of O2 produced.
From the balanced equation, 2 moles of KClO3 produce 3 moles of O2. Therefore, the number of moles of O2 produced from 0.0816 moles of KClO3 is:
\[ \text{Moles of O}_2 = 0.0816 \text{ mol KClO}_3 \times \frac{3 \text{ mol O}_2}{2 \text{ mol KClO}_3} = 0.1224 \text{ mol O}_2 \]

### Step 5: Convert the temperature from Celsius to Kelvin.
\[ T(K) = 18 + 273.15 = 291.15 \text{ K} \]

### Step 6: Convert the pressure from mmHg to atm.
Using the conversion factor \(1 \text{ atm} = 760 \text{ mmHg}\):
\[ P(\text{atm}) = \frac{750 \text{ mmHg}}{760 \text{ mmHg/atm}} \approx 0.9868 \text{ atm} \]

### Step 7: Use the ideal gas law to calculate the volume of O2.
The ideal gas law is given by:
\[ PV = nRT \]
Rearranging to solve for volume (V):
\[ V = \frac{nRT}{P} \]
Substituting the known values:
- \(n = 0.1224 \text{ mol}\)
- \(R = 0.0821 \text{ L atm/(K mol)}\)
- \(T = 291.15 \text{ K}\)
- \(P = 0.9868 \text{ atm}\)

Calculating the volume:
\[ V = \frac{0.1224 \text{ mol} \times 0.0821 \text{ L atm/(K mol)} \times 291.15 \text{ K}}{0.9868 \text{ atm}} \approx 2.96 \text{ L} \]

### Final Answer:
The volume of oxygen gas that can be obtained from 10 g of potassium chlorate at 18°C and 750 mmHg is approximately **2.96 liters**.

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