What is the minimum mass of `CaCO_3(s)`, below which it decomposes completely, required to establish equilibrium in a 6.50 litre container for the reaction :
`CaCO_3(s)hArr CaO(s)+CO_2(g), K_c=0.05`

Options

A. 32.5 g
B. 24.6 g
C. 40.6 g
D. 8.0 gm

Explanation

To solve the problem of determining the minimum mass of \( \text{CaCO}_3(s) \) required to establish equilibrium in a 6.50-liter container for the reaction:

\[ \text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \]

with \( K_c = 0.05 \), we can follow these steps:

### Step 1: Define the Variables
Let the minimum mass of calcium carbonate be \( x \) grams. The molar mass of \( \text{CaCO}_3 \) is approximately 100 g/mol.

### Step 2: Calculate Initial Concentration
The initial number of moles of \( \text{CaCO}_3 \) can be calculated as:
\[ \text{Moles of } \text{CaCO}_3 = \frac{x}{100} \]
The initial concentration of \( \text{CaCO}_3 \) in the 6.50 L container is:
\[ [\text{CaCO}_3] = \frac{\frac{x}{100}}{6.50} = \frac{x}{650} \]

### Step 3: Determine Equilibrium Concentrations
Since \( \text{CaCO}_3 \) decomposes completely, at equilibrium:
- The concentration of \( \text{CaCO}_3 \) will be 0.
- The concentration of \( \text{CaO} \) will be equal to the moles of \( \text{CaCO}_3 \) that decomposed, which is also \( \frac{x}{100} \) moles.
- The concentration of \( \text{CO}_2 \) will also be \( \frac{x}{100} \) moles.

Thus, the equilibrium concentrations can be expressed as:
\[ [\text{CaO}] = \frac{x}{100 \times 6.50} = \frac{x}{650} \]
\[ [\text{CO}_2] = \frac{x}{100 \times 6.50} = \frac{x}{650} \]

### Step 4: Write the Expression for \( K_c \)
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[\text{CO}_2]}{[\text{CaCO}_3]} \]
Since \( [\text{CaCO}_3] = 0 \) at equilibrium, we only consider the gaseous products:
\[ K_c = \frac{[\text{CO}_2]}{1} = [\text{CO}_2] \]
Substituting the expression for \( [\text{CO}_2] \):
\[ 0.05 = \frac{x}{650} \]

### Step 5: Solve for \( x \)
Rearranging the equation to solve for \( x \):
\[ x = 0.05 \times 650 \]
\[ x = 32.5 \text{ grams} \]

### Conclusion
The minimum mass of \( \text{CaCO}_3(s) \) required to establish equilibrium in the 6.50-liter container is **32.5 grams**.

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