1. Let the balanced reaction be:

$aBr_2 + bNa_2CO_3 \rightarrow cNaBr + dNaBrO_3 + eCO_2$

Since the reaction is in aqueous solution, water may also participate. So we write the more complete form:

$aBr_2 + bNa_2CO_3 + fH_2O \rightarrow cNaBr + dNaBrO_3 + eCO_2 + gH^+ \text{ / or equivalent balancing in basic medium}$

But a simpler way is to use the known disproportionation of bromine in alkaline medium.

2. In alkaline medium, bromine disproportionates as:

$3Br_2 + 6OH^- \rightarrow 5Br^- + BrO_3^- + 3H_2O$

This shows that for every $3$ molecules of $Br_2$, we get:

• $5$ bromide ions
• $1$ bromate ion

So the sodium salts formed will be:

$3Br_2 + 6NaOH \rightarrow 5NaBr + NaBrO_3 + 3H_2O$

3. Now replace $NaOH$ by $Na_2CO_3$. Carbonate provides the basic medium, and $CO_2$ is evolved. The balanced molecular equation is:

$3Br_2 + 3Na_2CO_3 \rightarrow 5NaBr + NaBrO_3 + 3CO_2$

Let us verify atom balance:

• Bromine: Left $= 3 \times 2 = 6$, Right $= 5 + 1 = 6$
• Sodium: Left $= 3 \times 2 = 6$, Right $= 5 + 1 = 6$
• Carbon: Left $= 3$, Right $= 3$
• Oxygen: Left $= 3 \times 3 = 9$, Right $= 3$ (in $NaBrO_3$) $+ 3 \times 2 = 6$ (in $CO_2$), total $9$

So it is balanced.

4. Therefore, the number of sodium bromide molecules involved is:

$\boxed{5}$