Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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![### Chemical Reaction and Carbon Dioxide Volume Calculation
#### Problem Statement
What volume of carbon dioxide is produced when 16.1 g of calcium carbonate reacts completely according to the following reaction at 25°C and 1 atm?
#### Chemical Reaction
\[ \text{calcium carbonate (s)} \rightarrow \text{calcium oxide (s)} + \text{carbon dioxide(g)} \]
#### Data Input
* Mass of calcium carbonate: 16.1 g
* Temperature: 25°C
* Pressure: 1 atm
#### Data Required
* Volume of carbon dioxide produced (in liters)
#### Procedure
1. **Identify the molar mass of calcium carbonate (CaCO₃)**:
- Calcium (Ca): 40.08 g/mol
- Carbon (C): 12.01 g/mol
- Oxygen (O): 16.00 g/mol (x 3)
- Molar mass of CaCO₃ = 40.08 + 12.01 + (16.00 x 3) = 100.09 g/mol
2. **Calculate the moles of calcium carbonate**:
\[
\text{Moles of CaCO₃} = \frac{\text{mass of CaCO₃}}{\text{molar mass of CaCO₃}} = \frac{16.1\ g}{100.09\ g/mol}
\]
3. **Use the stoichiometry of the reaction**:
- The reaction shows a 1:1 molar ratio between CaCO₃ and CO₂.
- Therefore, moles of CaCO₃ will equal moles of CO₂ produced.
4. **Calculate the volume of CO₂ produced**:
- Use the Ideal Gas Law: PV = nRT
- Where:
- P is the pressure (1 atm),
- V is the volume,
- n is the number of moles,
- R is the gas constant (0.0821 L atm / K mol),
- T is the temperature in Kelvin (25°C + 273.15 = 298.15 K).
\[
V = \frac{nRT}{P} = \frac{(\text{moles of CO₂}) \cdot (0.0821\ L\ atm / K\ mol) \cdot (298.](https://content.bartleby.com/qna-images/question/60358686-21d1-41ff-a93a-9ba59c7b6516/1ddd1e70-c5e8-4bff-b6af-e6513a43ab60/73r33i_reoriented.jpeg)
Transcribed Image Text:### Chemical Reaction and Carbon Dioxide Volume Calculation
#### Problem Statement
What volume of carbon dioxide is produced when 16.1 g of calcium carbonate reacts completely according to the following reaction at 25°C and 1 atm?
#### Chemical Reaction
\[ \text{calcium carbonate (s)} \rightarrow \text{calcium oxide (s)} + \text{carbon dioxide(g)} \]
#### Data Input
* Mass of calcium carbonate: 16.1 g
* Temperature: 25°C
* Pressure: 1 atm
#### Data Required
* Volume of carbon dioxide produced (in liters)
#### Procedure
1. **Identify the molar mass of calcium carbonate (CaCO₃)**:
- Calcium (Ca): 40.08 g/mol
- Carbon (C): 12.01 g/mol
- Oxygen (O): 16.00 g/mol (x 3)
- Molar mass of CaCO₃ = 40.08 + 12.01 + (16.00 x 3) = 100.09 g/mol
2. **Calculate the moles of calcium carbonate**:
\[
\text{Moles of CaCO₃} = \frac{\text{mass of CaCO₃}}{\text{molar mass of CaCO₃}} = \frac{16.1\ g}{100.09\ g/mol}
\]
3. **Use the stoichiometry of the reaction**:
- The reaction shows a 1:1 molar ratio between CaCO₃ and CO₂.
- Therefore, moles of CaCO₃ will equal moles of CO₂ produced.
4. **Calculate the volume of CO₂ produced**:
- Use the Ideal Gas Law: PV = nRT
- Where:
- P is the pressure (1 atm),
- V is the volume,
- n is the number of moles,
- R is the gas constant (0.0821 L atm / K mol),
- T is the temperature in Kelvin (25°C + 273.15 = 298.15 K).
\[
V = \frac{nRT}{P} = \frac{(\text{moles of CO₂}) \cdot (0.0821\ L\ atm / K\ mol) \cdot (298.
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