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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
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