To solve the problem step by step, we will follow the outlined approach in the video transcript.
### Step 1: Identify the Reaction and Given Data
The reaction is:
\[ \text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \]
Given data:
- \( K_p = 1.16 \, \text{atm} \)
- Mass of \( \text{CaCO}_3 = 20 \, \text{g} \)
- Volume of the vessel = 10 L
- Temperature = \( 800^\circ C = 1073 \, \text{K} \)
### Step 2: Calculate the Moles of CaCO₃
First, we need to calculate the number of moles of \( \text{CaCO}_3 \) present initially.
The molar mass of \( \text{CaCO}_3 \) is approximately:
- Ca: 40 g/mol
- C: 12 g/mol
- O: 16 g/mol × 3 = 48 g/mol
- Total = 100 g/mol
Now, calculate the moles:
\[
\text{Moles of } \text{CaCO}_3 = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{20 \, \text{g}}{100 \, \text{g/mol}} = 0.2 \, \text{mol}
\]
### Step 3: Determine the Partial Pressure of CO₂
According to the equilibrium expression, \( K_p \) is related to the partial pressure of \( \text{CO}_2 \):
\[
K_p = P_{\text{CO}_2} = 1.16 \, \text{atm}
\]
### Step 4: Use the Ideal Gas Law to Find the Mass of CO₂
We can use the ideal gas law to find the mass of \( \text{CO}_2 \) produced at equilibrium:
\[
PV = nRT \quad \Rightarrow \quad n = \frac{PV}{RT}
\]
Where:
- \( P = 1.16 \, \text{atm} \)
- \( V = 10 \, \text{L} \)
- \( R = 0.0821 \, \text{L atm/(mol K)} \)
- \( T = 1073 \, \text{K} \)
Substituting the values:
\[
n = \frac{(1.16 \, \text{atm})(10 \, \text{L})}{(0.0821 \, \text{L atm/(mol K)})(1073 \, \text{K})}
\]
Calculating:
\[
n = \frac{11.6}{88.0363} \approx 0.1316 \, \text{mol}
\]
Now, calculate the mass of \( \text{CO}_2 \):
\[
\text{Mass of } \text{CO}_2 = n \times \text{Molar Mass of } \text{CO}_2 = 0.1316 \, \text{mol} \times 44 \, \text{g/mol} \approx 5.8 \, \text{g}
\]
### Step 5: Calculate the Mass of CaCO₃ that Reacted
From the stoichiometry of the reaction, 1 mole of \( \text{CaCO}_3 \) produces 1 mole of \( \text{CO}_2 \). Thus, the mass of \( \text{CaCO}_3 \) that reacted can be calculated as follows:
Let \( x \) be the mass of \( \text{CaCO}_3 \) that reacted:
\[
\frac{x}{100} = \frac{5.8}{44}
\]
Calculating \( x \):
\[
x = \frac{5.8 \times 100}{44} \approx 13.18 \, \text{g}
\]
### Step 6: Calculate the Mass of Unreacted CaCO₃
The mass of unreacted \( \text{CaCO}_3 \) is:
\[
\text{Mass of unreacted } \text{CaCO}_3 = 20 \, \text{g} - 13.18 \, \text{g} \approx 6.82 \, \text{g}
\]
### Step 7: Conclusion
The amount of \( \text{CaCO}_3 \) remaining at equilibrium is approximately \( 6.82 \, \text{g} \).
