Reaction of `Br_(2)` with `Na_(2)CO_(3)` in aqueous solution gives NaBr and sodium bormate with evolution of `CO_(2)` gas. The number of NaBr molecue involved in the balanced chemical equation is ………….

Explanation

To solve the problem, we need to write and balance the chemical equation for the reaction of bromine (\(Br_2\)) with sodium carbonate (\(Na_2CO_3\)) in an aqueous solution. The products of this reaction are sodium bromide (\(NaBr\)), sodium bromate (\(NaBrO_3\)), and carbon dioxide (\(CO_2\)).

### Step-by-Step Solution:

1. **Write the unbalanced equation**:
The reactants are bromine and sodium carbonate, and the products are sodium bromide, sodium bromate, and carbon dioxide.
\[ Br_2 + Na_2CO_3 \rightarrow NaBr + NaBrO_3 + CO_2 \]

2. **Identify the number of moles of each reactant and product**:
From the information given, we know that:
- 3 moles of \(Na_2CO_3\) react with 3 moles of \(Br_2\).
- The products formed are 5 moles of \(NaBr\) and 3 moles of \(CO_2\).

3. **Balance the equation**:
Based on the stoichiometry provided:
\[ 3 Na_2CO_3 + 3 Br_2 \rightarrow 5 NaBr + 3 NaBrO_3 + 3 CO_2 \]

4. **Count the number of \(NaBr\) molecules**:
From the balanced equation, we see that 5 moles of sodium bromide (\(NaBr\)) are produced.

5. **Final answer**:
The number of \(NaBr\) molecules involved in the balanced chemical equation is 5.

### Summary:
The balanced chemical equation shows that 5 moles of sodium bromide are produced when bromine reacts with sodium carbonate.