`CH_(3)-overset(Cl)overset(|)underset(Cl)underset(|)(C)-CH_(3)` to `CH_(3)-C-=CNa` conversion by using `NaNH_(2)//Liq.NH_(3)`, Number of moles of `NaNH_(2)` required will be

Explanation

To convert the compound \( CH_3-CHCl-CHCl-CH_3 \) to \( CH_3-C \equiv CNa \) using \( NaNH_2 \) in liquid ammonia, we need to analyze the reaction step by step.

### Step 1: Identify the Starting Compound
The starting compound is \( CH_3-CHCl-CHCl-CH_3 \). It has two chlorine atoms attached to the second carbon.

### Step 2: First Reaction with \( NaNH_2 \)
In the first step, we will use one mole of \( NaNH_2 \) to remove one of the chlorine atoms and one hydrogen atom from the adjacent carbon. This will result in the formation of a double bond (alkene) and the release of \( HCl \).

**Reaction:**
\[ CH_3-CHCl-CHCl-CH_3 + NaNH_2 \rightarrow CH_3-CH=CH-CH_3 + HCl \]

### Step 3: Second Reaction with \( NaNH_2 \)
Next, we will use another mole of \( NaNH_2 \) to remove the second chlorine atom and a hydrogen atom from the adjacent carbon. This will convert the alkene into an alkyne.

**Reaction:**
\[ CH_3-CH=CH-CH_3 + NaNH_2 \rightarrow CH_3-C \equiv C-CH_3 + HCl \]

### Step 4: Final Reaction to Form the Desired Product
Finally, we will use one more mole of \( NaNH_2 \) to generate the sodium salt of the alkyne.

**Reaction:**
\[ CH_3-C \equiv C-CH_3 + NaNH_2 \rightarrow CH_3-C \equiv CNa + NH_3 \]

### Conclusion
In total, we used 3 moles of \( NaNH_2 \) to convert \( CH_3-CHCl-CHCl-CH_3 \) to \( CH_3-C \equiv CNa \).

### Final Answer
The number of moles of \( NaNH_2 \) required is **3 moles**.

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