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To solve this problem and find the percentage yield of the potassium chlorate decomposition reaction, let’s proceed step-by-step:
### Step 1: Identify the given values
- Actual volume of oxygen formed at room temperature: [tex]\( V_1 = 2.9 \, \text{dm}^3 \)[/tex]
- Room temperature: [tex]\( 25^\circ \text{C} \)[/tex]
- Temperature at Standard Temperature and Pressure (STP): [tex]\( 0^\circ \text{C} \)[/tex]
- Standard pressure: [tex]\( 1 \, \text{atm} \)[/tex]
### Step 2: Convert temperatures to Kelvin
Temperature must be in Kelvin for gas law calculations.
- Room temperature in Kelvin:
[tex]\[ T_1 = 25^\circ \text{C} + 273.15 = 298.15 \, \text{K} \][/tex]
- STP temperature in Kelvin:
[tex]\[ T_2 = 0^\circ \text{C} + 273.15 = 273.15 \, \text{K} \][/tex]
### Step 3: Use the ideal gas law relation
The ideal gas law states that for a given amount of gas at constant pressure:
[tex]\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \][/tex]
Where:
- [tex]\( V_1 \)[/tex] is the actual volume at room temperature
- [tex]\( T_1 \)[/tex] is the room temperature in Kelvin
- [tex]\( V_2 \)[/tex] is the theoretical volume at STP
- [tex]\( T_2 \)[/tex] is the temperature at STP in Kelvin
Rearranging to solve for [tex]\( V_2 \)[/tex]:
[tex]\[ V_2 = V_1 \cdot \frac{T_2}{T_1} \][/tex]
### Step 4: Calculate the theoretical volume at STP
Substituting the known values into the equation:
[tex]\[ V_2 = 2.9 \, \text{dm}^3 \cdot \frac{273.15 \, \text{K}}{298.15 \, \text{K}} = 2.6568338084856613 \, \text{dm}^3 \][/tex]
### Step 5: Calculate the percentage yield
Using the formula for percentage yield:
[tex]\[ \text{Percentage Yield} = \left( \frac{\text{Actual Volume}}{\text{Theoretical Volume}} \right) \times 100 \][/tex]
Substitute the actual and theoretical volumes:
[tex]\[ \text{Percentage Yield} = \left( \frac{2.9 \, \text{dm}^3}{2.6568338084856613 \, \text{dm}^3} \right) \times 100 \approx 109.1524803221673 \% \][/tex]
### Conclusion
Thus, the percentage yield for the potassium chlorate decomposition reaction is approximately [tex]\( 109.15\% \)[/tex].
### Step 1: Identify the given values
- Actual volume of oxygen formed at room temperature: [tex]\( V_1 = 2.9 \, \text{dm}^3 \)[/tex]
- Room temperature: [tex]\( 25^\circ \text{C} \)[/tex]
- Temperature at Standard Temperature and Pressure (STP): [tex]\( 0^\circ \text{C} \)[/tex]
- Standard pressure: [tex]\( 1 \, \text{atm} \)[/tex]
### Step 2: Convert temperatures to Kelvin
Temperature must be in Kelvin for gas law calculations.
- Room temperature in Kelvin:
[tex]\[ T_1 = 25^\circ \text{C} + 273.15 = 298.15 \, \text{K} \][/tex]
- STP temperature in Kelvin:
[tex]\[ T_2 = 0^\circ \text{C} + 273.15 = 273.15 \, \text{K} \][/tex]
### Step 3: Use the ideal gas law relation
The ideal gas law states that for a given amount of gas at constant pressure:
[tex]\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \][/tex]
Where:
- [tex]\( V_1 \)[/tex] is the actual volume at room temperature
- [tex]\( T_1 \)[/tex] is the room temperature in Kelvin
- [tex]\( V_2 \)[/tex] is the theoretical volume at STP
- [tex]\( T_2 \)[/tex] is the temperature at STP in Kelvin
Rearranging to solve for [tex]\( V_2 \)[/tex]:
[tex]\[ V_2 = V_1 \cdot \frac{T_2}{T_1} \][/tex]
### Step 4: Calculate the theoretical volume at STP
Substituting the known values into the equation:
[tex]\[ V_2 = 2.9 \, \text{dm}^3 \cdot \frac{273.15 \, \text{K}}{298.15 \, \text{K}} = 2.6568338084856613 \, \text{dm}^3 \][/tex]
### Step 5: Calculate the percentage yield
Using the formula for percentage yield:
[tex]\[ \text{Percentage Yield} = \left( \frac{\text{Actual Volume}}{\text{Theoretical Volume}} \right) \times 100 \][/tex]
Substitute the actual and theoretical volumes:
[tex]\[ \text{Percentage Yield} = \left( \frac{2.9 \, \text{dm}^3}{2.6568338084856613 \, \text{dm}^3} \right) \times 100 \approx 109.1524803221673 \% \][/tex]
### Conclusion
Thus, the percentage yield for the potassium chlorate decomposition reaction is approximately [tex]\( 109.15\% \)[/tex].
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