`K_(p)` for `CaCO_(3)(s) hArrCaO(s)+CO_(2)(g)` is `0.5` at `1000K ` 2 moles of `CaO(s) & CO_(2)(g)` each at `0.45 atm` introduce in a `16.4` lit. vessel and heated upto `1000 K`. The amount of `CaCO_(3)(s)` formed will be.

Options

A. 2 mole
B. `0.01` mole
C. `1.9` mole
D. 1 mole

Explanation

To solve the problem, we need to determine the amount of calcium carbonate (CaCO₃) formed when 2 moles each of calcium oxide (CaO) and carbon dioxide (CO₂) are introduced into a 16.4 L vessel and heated to 1000 K. The equilibrium constant \( K_p \) for the reaction is given as 0.5 at this temperature.

### Step-by-Step Solution

1. **Write the equilibrium reaction:**
\[ \text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \]
Here, only CO₂ is in the gaseous state, while CaCO₃ and CaO are solids.

2. **Identify the expression for \( K_p \):**
Since \( K_p \) is only dependent on the gaseous components, we have:
\[ K_p = \frac{P_{\text{CO}_2}}{1} = P_{\text{CO}_2} \]
Given \( K_p = 0.5 \), we have:
\[ P_{\text{CO}_2} = 0.5 \text{ atm} \]

3. **Calculate the number of moles of CO₂ at equilibrium:**
Using the ideal gas law:
\[ PV = nRT \]
Rearranging gives:
\[ n = \frac{PV}{RT} \]
Where:
- \( P = 0.5 \) atm
- \( V = 16.4 \) L
- \( R = 0.082 \) L·atm/(K·mol)
- \( T = 1000 \) K

Plugging in the values:
\[ n = \frac{0.5 \times 16.4}{0.082 \times 1000} \]
\[ n = \frac{8.2}{82} = 0.1 \text{ moles of CO}_2 \]

4. **Set up the equilibrium expression:**
Initially, we have:
- Moles of CaO = 2
- Moles of CO₂ = 2

At equilibrium, let \( x \) be the moles of CaCO₃ formed. Thus:
- Moles of CaO left = \( 2 - x \)
- Moles of CO₂ left = \( 2 - x \)

From the equilibrium condition:
\[ P_{\text{CO}_2} = (2 - x) \text{ atm} \]
Setting this equal to the equilibrium pressure:
\[ 2 - x = 0.1 \]

5. **Solve for \( x \):**
\[ x = 2 - 0.1 = 1.9 \]
Therefore, the amount of CaCO₃ formed is 1.9 moles.

### Final Answer
The amount of calcium carbonate (CaCO₃) formed is **1.9 moles**.