`2.616g` of an element (X) is heated with excess of `NaNO_(3)` and `NaOH` to produce a basic gas (Y) and `Na_(2)XO_(2)`. The basic gas liberated exactly requires `5 xx 10^(-3)` mol `H_(2)SO_(4)` for complete neutralisation. If 4 mol of X are completely consumed by caustic soda, find the no of mole of `H_(2)` gas liberated.

Explanation

To solve the problem step by step, we will analyze the information provided and perform the necessary calculations.

### Step 1: Understanding the Reaction
The element \( X \) reacts with \( NaNO_3 \) and \( NaOH \) to produce a basic gas \( Y \) and \( Na_2XO_2 \). The basic gas \( Y \) is neutralized by \( H_2SO_4 \).

### Step 2: Determine Moles of \( H_2SO_4 \)
We know that the basic gas \( Y \) requires \( 5 \times 10^{-3} \) moles of \( H_2SO_4 \) for complete neutralization.

### Step 3: Relate Moles of \( X \) to Moles of \( H_2 \)
The problem states that 4 moles of \( X \) are consumed. We need to find out how many moles of \( H_2 \) gas are liberated in the reaction.

### Step 4: Write the Reaction
The balanced reaction for the formation of \( Na_2XO_2 \) and the liberation of \( H_2 \) can be represented as:
\[ 4X + 4NaOH \rightarrow Na_2XO_2 + 2H_2 + 2NH_3 + 2H_2O \]
From the reaction, we can see that for every 4 moles of \( X \), 2 moles of \( H_2 \) are produced.

### Step 5: Calculate Moles of \( H_2 \)
From the stoichiometry of the reaction:
- If 4 moles of \( X \) yield 2 moles of \( H_2 \), then:
\[ \text{Moles of } H_2 = \frac{2 \text{ moles of } H_2}{4 \text{ moles of } X} \times 4 \text{ moles of } X = 2 \text{ moles of } H_2 \]

### Step 6: Conclusion
The number of moles of \( H_2 \) gas liberated is **2 moles**.

### Final Answer
The number of moles of \( H_2 \) gas liberated is **2 moles**.

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