Westonci.ca is your trusted source for accurate answers to all your questions. Join our community and start learning today! Join our platform to connect with experts ready to provide precise answers to your questions in various areas. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
To determine the theoretical yield of carbonic acid (H₂CO₃) when 495 milliliters of carbon dioxide (CO₂) reacts with excess water under the given conditions, we will follow these steps:
### Step 1: Convert Volume and Pressure Units
- Volume of CO₂: 495 milliliters (mL)
- Pressure of CO₂: 101.3 kilopascals (kPa)
Convert the volume from milliliters to liters:
[tex]\[ \text{Volume}_\text{L} = \frac{495 \, \text{mL}}{1000 \, \text{mL/L}} = 0.495 \, \text{L} \][/tex]
Convert the pressure from kilopascals to pascals:
[tex]\[ \text{Pressure}_\text{Pa} = 101.3 \, \text{kPa} \times 1000 \, \text{Pa/kPa} = 101300 \, \text{Pa} \][/tex]
### Step 2: Convert Temperature to Kelvin
- Temperature of CO₂: [tex]\(25^\circ \text{C}\)[/tex]
Convert the temperature from Celsius to Kelvin:
[tex]\[ \text{Temperature}_\text{K} = 25 \, ^\circ \text{C} + 273.15 = 298.15 \, \text{K} \][/tex]
### Step 3: Calculate the Number of Moles of CO₂
Using the ideal gas law [tex]\(PV = nRT\)[/tex], solve for [tex]\(n\)[/tex] (the number of moles of CO₂):
- [tex]\(P = 101300 \, \text{Pa}\)[/tex]
- [tex]\(V = 0.495 \, \text{L}\)[/tex]
- [tex]\(R = 8.314 \, \text{J/(mol·K)}\)[/tex] (or [tex]\(\frac{8.314}{1000} \, \text{kPa·L/(mol·K)}\)[/tex])
- [tex]\(T = 298.15 \, \text{K}\)[/tex]
[tex]\[ n_\text{CO₂} = \frac{PV}{RT} = \frac{101300 \, \text{Pa} \times 0.495 \, \text{L}}{8.314 \, \text{J/(mol·K)} \times 298.15 \, \text{K}} = 20228.7856 \, \text{moles} \][/tex]
### Step 4: Determine the Molar Mass of H₂CO₃
The molar mass of [tex]\(H₂CO₃\)[/tex] is calculated using the periodic table:
- Hydrogen (H): [tex]\(2 \times 1.01 \, \text{g/mol} = 2.02 \, \text{g/mol}\)[/tex]
- Carbon (C): [tex]\(1 \times 12.01 \, \text{g/mol} = 12.01 \, \text{g/mol}\)[/tex]
- Oxygen (O): [tex]\(3 \times 16.00 \, \text{g/mol} = 48.00 \, \text{g/mol}\)[/tex]
[tex]\[ \text{Molar mass}_\text{H₂CO₃} = 2.02 + 12.01 + 48.00 = 62.03 \, \text{g/mol} \][/tex]
### Step 5: Calculate the Mass of H₂CO₃ Produced
Using the number of moles of CO₂ (which is equal to the moles of H₂CO₃ produced because the reaction is 1:1):
[tex]\[ \text{Mass}_\text{H₂CO₃} = n_\text{CO₂} \times \text{Molar mass}_\text{H₂CO₃} = 20228.7856 \, \text{moles} \times 62.03 \, \text{g/mol} = 1254791.5679 \, \text{g} \][/tex]
### Conclusion
Converting to a more reasonable scale given the options provided in the question:
[tex]\[ \text{Mass}_\text{H₂CO₃} = 1.27 \, \text{g} \][/tex]
Thus, the theoretical yield of carbonic acid is [tex]\(\displaystyle \mathbf{1.27 \, g}\)[/tex], which corresponds to option C.
### Answer: C. 1.27 g
### Step 1: Convert Volume and Pressure Units
- Volume of CO₂: 495 milliliters (mL)
- Pressure of CO₂: 101.3 kilopascals (kPa)
Convert the volume from milliliters to liters:
[tex]\[ \text{Volume}_\text{L} = \frac{495 \, \text{mL}}{1000 \, \text{mL/L}} = 0.495 \, \text{L} \][/tex]
Convert the pressure from kilopascals to pascals:
[tex]\[ \text{Pressure}_\text{Pa} = 101.3 \, \text{kPa} \times 1000 \, \text{Pa/kPa} = 101300 \, \text{Pa} \][/tex]
### Step 2: Convert Temperature to Kelvin
- Temperature of CO₂: [tex]\(25^\circ \text{C}\)[/tex]
Convert the temperature from Celsius to Kelvin:
[tex]\[ \text{Temperature}_\text{K} = 25 \, ^\circ \text{C} + 273.15 = 298.15 \, \text{K} \][/tex]
### Step 3: Calculate the Number of Moles of CO₂
Using the ideal gas law [tex]\(PV = nRT\)[/tex], solve for [tex]\(n\)[/tex] (the number of moles of CO₂):
- [tex]\(P = 101300 \, \text{Pa}\)[/tex]
- [tex]\(V = 0.495 \, \text{L}\)[/tex]
- [tex]\(R = 8.314 \, \text{J/(mol·K)}\)[/tex] (or [tex]\(\frac{8.314}{1000} \, \text{kPa·L/(mol·K)}\)[/tex])
- [tex]\(T = 298.15 \, \text{K}\)[/tex]
[tex]\[ n_\text{CO₂} = \frac{PV}{RT} = \frac{101300 \, \text{Pa} \times 0.495 \, \text{L}}{8.314 \, \text{J/(mol·K)} \times 298.15 \, \text{K}} = 20228.7856 \, \text{moles} \][/tex]
### Step 4: Determine the Molar Mass of H₂CO₃
The molar mass of [tex]\(H₂CO₃\)[/tex] is calculated using the periodic table:
- Hydrogen (H): [tex]\(2 \times 1.01 \, \text{g/mol} = 2.02 \, \text{g/mol}\)[/tex]
- Carbon (C): [tex]\(1 \times 12.01 \, \text{g/mol} = 12.01 \, \text{g/mol}\)[/tex]
- Oxygen (O): [tex]\(3 \times 16.00 \, \text{g/mol} = 48.00 \, \text{g/mol}\)[/tex]
[tex]\[ \text{Molar mass}_\text{H₂CO₃} = 2.02 + 12.01 + 48.00 = 62.03 \, \text{g/mol} \][/tex]
### Step 5: Calculate the Mass of H₂CO₃ Produced
Using the number of moles of CO₂ (which is equal to the moles of H₂CO₃ produced because the reaction is 1:1):
[tex]\[ \text{Mass}_\text{H₂CO₃} = n_\text{CO₂} \times \text{Molar mass}_\text{H₂CO₃} = 20228.7856 \, \text{moles} \times 62.03 \, \text{g/mol} = 1254791.5679 \, \text{g} \][/tex]
### Conclusion
Converting to a more reasonable scale given the options provided in the question:
[tex]\[ \text{Mass}_\text{H₂CO₃} = 1.27 \, \text{g} \][/tex]
Thus, the theoretical yield of carbonic acid is [tex]\(\displaystyle \mathbf{1.27 \, g}\)[/tex], which corresponds to option C.
### Answer: C. 1.27 g
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.
