For the reaction at equilibrium
`CaCO_(3) hArr CaO(s) + CO_(2)(g) `
What `CaO(s)` is removed reaction moves in forward direction.

Answer

To determine whether the statement "If `CaO(s)` is removed, the reaction moves in the forward direction" is true or false, we can analyze the equilibrium reaction:

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

### Step-by-Step Solution:

1. **Understand the Equilibrium Concept:**
- At equilibrium, the rates of the forward and reverse reactions are equal, and the concentrations of the reactants and products remain constant.

2. **Identify the Components:**
- In the given reaction, `CaCO3` is a solid reactant, `CaO` is a solid product, and `CO2` is a gaseous product.

3. **Apply Le Chatelier's Principle:**
- Le Chatelier's Principle states that if a system at equilibrium is subjected to a change (such as concentration, pressure, or temperature), the system will adjust to counteract that change and restore a new equilibrium.

4. **Consider the Effect of Removing `CaO(s)`:**
- According to Le Chatelier's Principle, if a product is removed from the equilibrium mixture, the system will respond by shifting the equilibrium to the right (forward direction) to produce more of that product.

5. **Evaluate the Role of `CaO(s)`:**
- However, `CaO` is a solid. In equilibrium expressions, the concentrations of pure solids and liquids do not change and are not included in the equilibrium constant expression. Therefore, removing `CaO(s)` does not affect the equilibrium position.

6. **Conclusion:**
- Since the removal of a solid does not shift the equilibrium, the statement "If `CaO(s)` is removed, the reaction moves in the forward direction" is **false**.

### Final Answer:
The statement is **false**.

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