Urea `(H_(2)NCONH_(2))+` hot dilute NaOH `to X+NH_(3), X ` is

Options

A. `NaNH-underset(O)underset(||)C-NH.Na`
B. `NH_(2)-underset(O)underset(||)C-NH-Na`
C. `Na_(2)CO_(3)`
D. `H-underset(O)underset(||)C-NH_(2)`

Answer

To solve the question regarding the reaction of urea with hot dilute NaOH, we can follow these steps:

### Step 1: Write the chemical equation for the reaction
Urea (H₂NCONH₂) reacts with hot dilute sodium hydroxide (NaOH). The reaction can be represented as follows:

\[ \text{H}_2\text{NCONH}_2 + \text{NaOH} \rightarrow X + \text{NH}_3 \]

### Step 2: Identify the products
When urea reacts with hot dilute NaOH, it undergoes hydrolysis, leading to the formation of sodium carbonate (Na₂CO₃) and ammonia (NH₃). Therefore, we can express the reaction as:

\[ \text{H}_2\text{NCONH}_2 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{CO}_3 + 2 \text{NH}_3 \]

### Step 3: Determine the value of X
From the balanced equation, we can see that X is sodium carbonate (Na₂CO₃).

### Final Answer
Thus, the product X formed when urea reacts with hot dilute NaOH is:

\[ X = \text{Na}_2\text{CO}_3 \]

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