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To find the theoretical yield of carbonic acid ([tex]\(H_2CO_3\)[/tex]) formed when carbon dioxide ([tex]\(CO_2\)[/tex]) reacts with water, we'll use the ideal gas law and stoichiometry. Here is a step-by-step solution:
### Step 1: Data and Assumptions
1. Volume of [tex]\(CO_2\)[/tex]: [tex]\(495 \text{ milliliters}\)[/tex] (or [tex]\(0.495\)[/tex] liters)
2. Temperature: [tex]\(25^{\circ} \text{C}\)[/tex] (convert to Kelvin: [tex]\(25 + 273.15 = 298.15 \, \text{K}\)[/tex])
3. Pressure: [tex]\(101.3 \, \text{kPa}\)[/tex] (standard atmospheric pressure)
### Step 2: Ideal Gas Law
Use the ideal gas law [tex]\(PV = nRT\)[/tex] where:
- [tex]\(P\)[/tex] is pressure ([tex]\(101.3 \, \text{kPa}\)[/tex])
- [tex]\(V\)[/tex] is volume ([tex]\(0.495 \, \text{L}\)[/tex])
- [tex]\(n\)[/tex] is the number of moles of [tex]\(CO_2\)[/tex] we're looking for
- [tex]\(R\)[/tex] is the ideal gas constant in appropriate units: [tex]\( \frac{8.314 \, \text{L·kPa}}{\text{mol·K}}\)[/tex]
- [tex]\(T\)[/tex] is temperature in Kelvin ([tex]\(298.15 \, \text{K}\)[/tex])
Rearranging the formula to solve for [tex]\(n\)[/tex] (number of moles):
[tex]\[ n = \frac{PV}{RT} \][/tex]
### Step 3: Calculate Moles of [tex]\(CO_2\)[/tex]
Substitute the values:
[tex]\[ n_{CO_2} = \frac{101.3 \, kPa \times 0.495 \, L}{8.314 \, \frac{L·kPa}{mol·K} \times 298.15 \, K} \][/tex]
[tex]\[ n_{CO_2} = \frac{50.1035}{2477.2491} \][/tex]
[tex]\[ n_{CO_2} = 0.02049 \, \text{moles CO_2} \][/tex]
### Step 4: Reacting Ratios and Molar Mass
According to the chemical equation [tex]\(CO_2 + H_2O \rightarrow H_2CO_3\)[/tex], the mole ratio of [tex]\(CO_2\)[/tex] to [tex]\(H_2CO_3\)[/tex] is 1:1. Therefore, 0.02049 moles of [tex]\(CO_2\)[/tex] will produce 0.02049 moles of [tex]\(H_2CO_3\)[/tex].
Calculate the molar mass of [tex]\(H_2CO_3\)[/tex]:
[tex]\[ \text{Molar mass of H}_2\text{CO}_3 = 2 \times 1.01 + 12.01 + 3 \times 16.00 \][/tex]
[tex]\[ \text{Molar mass of H}_2\text{CO}_3 = 2.02 + 12.01 + 48.00 \][/tex]
[tex]\[ \text{Molar mass of H}_2\text{CO}_3 = 62.03 \, \text{g/mol} \][/tex]
### Step 5: Calculate Theoretical Yield of [tex]\(H_2CO_3\)[/tex]
Theoretical yield = moles of [tex]\(H_2CO_3 \times \)[/tex] molar mass of [tex]\(H_2CO_3\)[/tex]:
[tex]\[ \text{Theoretical yield} = 0.02049 \, \text{moles} \times 62.03 \, \text{g/mol} \][/tex]
[tex]\[ \text{Theoretical yield} = 1.27 \, \text{grams} \][/tex]
### Answer
The theoretical yield of carbonic acid is approximately [tex]\(1.27 \, \text{grams}\)[/tex]. Therefore, the correct answer is:
[tex]\[ \boxed{1.27 \, \text{g}} \][/tex]
Answer choice: C.
### Step 1: Data and Assumptions
1. Volume of [tex]\(CO_2\)[/tex]: [tex]\(495 \text{ milliliters}\)[/tex] (or [tex]\(0.495\)[/tex] liters)
2. Temperature: [tex]\(25^{\circ} \text{C}\)[/tex] (convert to Kelvin: [tex]\(25 + 273.15 = 298.15 \, \text{K}\)[/tex])
3. Pressure: [tex]\(101.3 \, \text{kPa}\)[/tex] (standard atmospheric pressure)
### Step 2: Ideal Gas Law
Use the ideal gas law [tex]\(PV = nRT\)[/tex] where:
- [tex]\(P\)[/tex] is pressure ([tex]\(101.3 \, \text{kPa}\)[/tex])
- [tex]\(V\)[/tex] is volume ([tex]\(0.495 \, \text{L}\)[/tex])
- [tex]\(n\)[/tex] is the number of moles of [tex]\(CO_2\)[/tex] we're looking for
- [tex]\(R\)[/tex] is the ideal gas constant in appropriate units: [tex]\( \frac{8.314 \, \text{L·kPa}}{\text{mol·K}}\)[/tex]
- [tex]\(T\)[/tex] is temperature in Kelvin ([tex]\(298.15 \, \text{K}\)[/tex])
Rearranging the formula to solve for [tex]\(n\)[/tex] (number of moles):
[tex]\[ n = \frac{PV}{RT} \][/tex]
### Step 3: Calculate Moles of [tex]\(CO_2\)[/tex]
Substitute the values:
[tex]\[ n_{CO_2} = \frac{101.3 \, kPa \times 0.495 \, L}{8.314 \, \frac{L·kPa}{mol·K} \times 298.15 \, K} \][/tex]
[tex]\[ n_{CO_2} = \frac{50.1035}{2477.2491} \][/tex]
[tex]\[ n_{CO_2} = 0.02049 \, \text{moles CO_2} \][/tex]
### Step 4: Reacting Ratios and Molar Mass
According to the chemical equation [tex]\(CO_2 + H_2O \rightarrow H_2CO_3\)[/tex], the mole ratio of [tex]\(CO_2\)[/tex] to [tex]\(H_2CO_3\)[/tex] is 1:1. Therefore, 0.02049 moles of [tex]\(CO_2\)[/tex] will produce 0.02049 moles of [tex]\(H_2CO_3\)[/tex].
Calculate the molar mass of [tex]\(H_2CO_3\)[/tex]:
[tex]\[ \text{Molar mass of H}_2\text{CO}_3 = 2 \times 1.01 + 12.01 + 3 \times 16.00 \][/tex]
[tex]\[ \text{Molar mass of H}_2\text{CO}_3 = 2.02 + 12.01 + 48.00 \][/tex]
[tex]\[ \text{Molar mass of H}_2\text{CO}_3 = 62.03 \, \text{g/mol} \][/tex]
### Step 5: Calculate Theoretical Yield of [tex]\(H_2CO_3\)[/tex]
Theoretical yield = moles of [tex]\(H_2CO_3 \times \)[/tex] molar mass of [tex]\(H_2CO_3\)[/tex]:
[tex]\[ \text{Theoretical yield} = 0.02049 \, \text{moles} \times 62.03 \, \text{g/mol} \][/tex]
[tex]\[ \text{Theoretical yield} = 1.27 \, \text{grams} \][/tex]
### Answer
The theoretical yield of carbonic acid is approximately [tex]\(1.27 \, \text{grams}\)[/tex]. Therefore, the correct answer is:
[tex]\[ \boxed{1.27 \, \text{g}} \][/tex]
Answer choice: C.
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