For the reaction ` : CaCO_(3)(s) hArr CaO(s)+CO_(2)(g),K_(p)=1.16atm` at `800^(@)C` . If `20g` of `CaCO_(3)` were kept in a 10 litre vessel at `800^(@)C`, the amount of `CaCO_(3)` remained at equilibrium is `:`

Options

A. `34%`
B. `64%`
C. `46%`
D. none

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the moles of CaCO₃
We are given the mass of CaCO₃ as 20 grams and its molar mass is 100 g/mol.

\[ \text{Moles of CaCO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{20 \text{ g}}{100 \text{ g/mol}} = 0.2 \text{ moles} \]

### Step 2: Set up the equilibrium expression
The reaction is:

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

At equilibrium, let \( x \) be the number of moles of CO₂ produced. The moles of CaCO₃ remaining will be \( 0.2 - x \).

### Step 3: Write the expression for Kp
The equilibrium constant \( K_p \) is given by:

\[ K_p = \frac{P_{\text{CO}_2}}{1} = 1.16 \text{ atm} \]

Since CaO is a solid, it does not appear in the expression.

### Step 4: Relate the pressure to moles of CO₂
Using the ideal gas law, we can find the moles of CO₂ produced at equilibrium:

\[ PV = nRT \implies n = \frac{PV}{RT} \]

Where:
- \( P = 1.16 \text{ atm} \)
- \( V = 10 \text{ L} \)
- \( R = 0.0821 \text{ L atm/(K mol)} \)
- \( T = 800 + 273 = 1073 \text{ K} \)

Substituting the values:

\[ n = \frac{(1.16 \text{ atm})(10 \text{ L})}{(0.0821 \text{ L atm/(K mol)})(1073 \text{ K})} \]

Calculating this gives:

\[ n \approx 0.132 \text{ moles of CO}_2 \]

### Step 5: Determine the moles of CaCO₃ remaining
Since \( x = 0.132 \) moles of CO₂ are produced, the moles of CaCO₃ remaining will be:

\[ \text{Moles of CaCO}_3 \text{ remaining} = 0.2 - 0.132 = 0.068 \text{ moles} \]

### Step 6: Calculate the mass of CaCO₃ remaining
To find the mass of CaCO₃ remaining, we multiply the moles by the molar mass:

\[ \text{Mass of CaCO}_3 \text{ remaining} = 0.068 \text{ moles} \times 100 \text{ g/mol} = 6.8 \text{ g} \]

### Step 7: Calculate the mass percentage of CaCO₃ remaining
The mass percentage of CaCO₃ remaining is calculated as follows:

\[ \text{Mass percentage} = \left( \frac{\text{mass remaining}}{\text{initial mass}} \right) \times 100 = \left( \frac{6.8 \text{ g}}{20 \text{ g}} \right) \times 100 = 34\% \]

### Final Answer
The amount of CaCO₃ that remains at equilibrium is **6.8 g**, which corresponds to **34%** of the initial mass.

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