Reaction of `Br_(2)` with `Na_(2)CO_(3)` in aqueous solution gives sodium bromide and sodium bromate with evolution of gas. The number of sodium bromide molecules formed in the balanced chemical equation is :

Explanation

To solve the problem of determining the number of sodium bromide (NaBr) molecules formed when bromine (Br₂) reacts with sodium carbonate (Na₂CO₃) in an aqueous solution, we need to follow these steps:

### Step 1: Write the unbalanced chemical equation
The reaction can be represented as:
\[ \text{Br}_2 + \text{Na}_2\text{CO}_3 \rightarrow \text{NaBr} + \text{NaBrO}_3 + \text{CO}_2 \]

### Step 2: Identify the products
From the reaction, we see that sodium bromide (NaBr) and sodium bromate (NaBrO₃) are formed along with the evolution of carbon dioxide (CO₂) gas.

### Step 3: Write the half-reaction for bromine
In a basic medium, bromine (Br₂) can be split into two half-reactions:
1. Reduction to bromide ion (Br⁻)
2. Oxidation to bromate ion (BrO₃⁻)

### Step 4: Balance the half-reactions
For the reduction half-reaction:
\[ \text{Br}_2 + 2e^- \rightarrow 2\text{Br}^- \]

For the oxidation half-reaction:
\[ \text{Br}_2 + 6\text{H}_2\text{O} \rightarrow 2\text{BrO}_3^- + 12\text{H}^+ + 10e^- \]

### Step 5: Combine the half-reactions
To combine the half-reactions, we need to balance the electrons. We can multiply the reduction half-reaction by 5 to equalize the number of electrons:
\[ 5\text{Br}_2 + 6\text{H}_2\text{O} \rightarrow 10\text{Br}^- + 2\text{BrO}_3^- + 12\text{H}^+ + 10e^- \]

### Step 6: Add hydroxide ions to balance in basic medium
We add hydroxide ions (OH⁻) to both sides to neutralize the H⁺ ions:
\[ 5\text{Br}_2 + 6\text{H}_2\text{O} + 12\text{OH}^- \rightarrow 10\text{Br}^- + 2\text{BrO}_3^- + 12\text{H}_2\text{O} \]

### Step 7: Simplify the equation
After canceling out water molecules, we get:
\[ 5\text{Br}_2 + 12\text{OH}^- \rightarrow 10\text{Br}^- + 2\text{BrO}_3^- + 6\text{H}_2\text{O} \]

### Step 8: Identify the number of NaBr produced
From the balanced equation, we see that for every 5 moles of Br₂, 10 moles of NaBr are produced. Since sodium bromide (NaBr) is produced in a 1:1 ratio with bromide ions (Br⁻), we conclude that:
- The number of sodium bromide molecules formed is 10.

### Final Answer
The number of sodium bromide molecules formed in the balanced chemical equation is **10**.

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