To solve the problem, we need to determine the mole percent of \( XCO_3 \) that remains unreacted at equilibrium after heating in a 50-liter container at 727 °C. We start with the given reaction:
\[
XCO_3 (s) \rightleftharpoons XO (s) + CO_2 (g)
\]
### Step 1: Write the expression for \( K_p \)
Since \( K_p \) is given for the reaction, we note that it only involves the gaseous products. The expression for \( K_p \) is:
\[
K_p = \frac{P_{CO_2}}{1} = P_{CO_2}
\]
where \( P_{CO_2} \) is the partial pressure of carbon dioxide.
### Step 2: Substitute the value of \( K_p \)
Given that \( K_p = 1.64 \, \text{atm} \), we have:
\[
P_{CO_2} = 1.64 \, \text{atm}
\]
### Step 3: Calculate the number of moles of \( CO_2 \) produced
Using the ideal gas law, we can find the number of moles of \( CO_2 \):
\[
PV = nRT
\]
Where:
- \( P = 1.64 \, \text{atm} \)
- \( V = 50 \, \text{L} \)
- \( R = 0.0821 \, \text{atm L/(mol K)} \)
- \( T = 727 + 273 = 1000 \, \text{K} \)
Rearranging the ideal gas equation to solve for \( n \):
\[
n = \frac{PV}{RT}
\]
Substituting in the values:
\[
n = \frac{(1.64 \, \text{atm})(50 \, \text{L})}{(0.0821 \, \text{atm L/(mol K)})(1000 \, \text{K})}
\]
Calculating this gives:
\[
n = \frac{82}{82.1} \approx 1 \, \text{mol}
\]
### Step 4: Determine the moles of \( XCO_3 \) that decomposed
From the stoichiometry of the reaction, 1 mole of \( CO_2 \) is produced from the decomposition of 1 mole of \( XCO_3 \). Therefore, the moles of \( XCO_3 \) that decomposed is 1 mole.
### Step 5: Calculate the initial moles of \( XCO_3 \)
Initially, we started with 4 moles of \( XCO_3 \).
### Step 6: Calculate the moles of \( XCO_3 \) that remain unreacted
The moles of \( XCO_3 \) that remain unreacted is:
\[
\text{Unreacted moles} = \text{Initial moles} - \text{Decomposed moles} = 4 - 1 = 3 \, \text{moles}
\]
### Step 7: Calculate the mole percent of \( XCO_3 \) that remains unreacted
To find the mole percent of \( XCO_3 \) that remains unreacted:
\[
\text{Mole percent unreacted} = \left( \frac{\text{Unreacted moles}}{\text{Initial moles}} \right) \times 100 = \left( \frac{3}{4} \right) \times 100 = 75\%
\]
### Final Answer
The mole percent of \( XCO_3 \) that remains unreacted at equilibrium is **75%**.
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