This question involves balancing a redox reaction where bromine reacts with sodium carbonate to form sodium bromide, sodium bromate, and carbon dioxide gas. Let's break it down step by step.

Step 1: Write the Skeleton Reaction

The unbalanced chemical equation based on the given information is:

${\mathrm{Br}}_2+{\mathrm{Na}}_2{\mathrm{CO}}_3\to \mathrm{NaBr}+\mathrm{NaBrO}{3}_{\mathrm{}}+{\mathrm{CO}}_2$

However, this is incomplete. The reaction produces CO₂ gas, indicating that the carbonate is being decomposed. A more standard representation for this type of reaction (halogen with carbonate) is:

$X_2+{\mathrm{Na}}_2{\mathrm{CO}}_3\to \mathrm{NaX}+\mathrm{NaXO}{3}_{\mathrm{}}+{\mathrm{CO}}_2$ (where X is a halogen)

For bromine (X = Br), the unbalanced equation is:

${\mathrm{Br}}_2+{\mathrm{Na}}_2{\mathrm{CO}}_3\to \mathrm{NaBr}+\mathrm{NaBrO}{3}_{\mathrm{}}+{\mathrm{CO}}_2$

We must balance this redox reaction.

Step 2: Identify the Oxidation and Reduction Half-Reactions

In this reaction, bromine (${\mathrm{Br}}_2$) undergoes disproportionation. One bromine atom is reduced, and the other is oxidized.

• Reduction: Some bromine (Oxidation state 0) gains electrons to become bromide ion, Br⁻ (Oxidation state -1). ${\mathrm{Br}}_2+2e^{-}\to 2{\mathrm{Br}}^{-}$
• Oxidation: Some bromine (Oxidation state 0) loses electrons to become bromate ion, BrO₃⁻ (Oxidation state +5). ${\mathrm{Br}}_2+6H_{2}O\to 2\mathrm{BrO}{3}_{\mathrm{}}^{-}+12H^{+}+10e^{-}$

The carbonate (${\mathrm{CO}}_3{}^{\mathrm{2-}}$) acts as a base and provides the medium for the reaction. It decomposes to release CO₂.

Step 3: Balance the Redox Process

To balance the number of electrons lost and gained, find the LCM of the electrons involved in the half-reactions (2 and 10). The LCM is 10.

Multiply the reduction half-reaction by 5 and the oxidation half-reaction by 1 to equalize the electrons (10e⁻).

• Reduction (×5): $5{\mathrm{Br}}_2+10e^{-}\to 10{\mathrm{Br}}^{-}$
• Oxidation (×1): ${\mathrm{Br}}_2+6H_{2}O\to 2\mathrm{BrO}{3}_{\mathrm{}}^{-}+12H^{+}+10e^{-}$

Now, add these two half-reactions together:

$5{\mathrm{Br}}_2+{\mathrm{Br}}_2+6H_{2}O+10e^{-}\to 10{\mathrm{Br}}^{-}+2\mathrm{BrO}{3}_{\mathrm{}}^{-}+12H^{+}+10e^{-}$

Cancel the 10e⁻ from both sides and combine the bromine molecules:

$6{\mathrm{Br}}_2+6H_{2}O\to 10{\mathrm{Br}}^{-}+2\mathrm{BrO}{3}_{\mathrm{}}^{-}+12H^{+}$

We have H⁺ ions on the product side. In an aqueous basic solution (provided by Na₂CO₃), these will be neutralized. Sodium carbonate (${\mathrm{Na}}_2{\mathrm{CO}}_3$) provides CO₃²⁻ ions which react with H⁺ to form CO₂ and H₂O: ${\mathrm{CO}}_3{}^{\mathrm{2-}}+2H^{+}\to {\mathrm{CO}}_2+H_{2}O$

We have 12 H⁺ ions, so we need 6 CO₃²⁻ ions to neutralize them. This will produce 6 CO₂ molecules and 6 H₂O molecules.

Let's incorporate this into our equation. Add 6 CO₃²⁻ to both sides (which will come from 3 Na₂CO₃). The final balanced equation in molecular form is:

$3{\mathrm{Br}}_2+3{\mathrm{Na}}_2{\mathrm{CO}}_3\to 5\mathrm{NaBr}+\mathrm{NaBrO}{3}_{\mathrm{}}+3{\mathrm{CO}}_2$

Verification: Check the number of atoms on both sides. They are balanced.

Final Answer

From the balanced equation $3{\mathrm{Br}}_2+3{\mathrm{Na}}_2{\mathrm{CO}}_3\to 5\mathrm{NaBr}+\mathrm{NaBrO}{3}_{\mathrm{}}+3{\mathrm{CO}}_2$, we can see that 5 molecules of sodium bromide (NaBr) are produced.

Therefore, the number of sodium bromide molecules involved is 5.

Related Topics & Formulae

Disproportionation: A redox reaction in which a species is simultaneously oxidized and reduced.

Balancing Redox Reactions: The ion-electron method (half-reaction method) is crucial, especially in acidic or basic media.

Role of Carbonate: Carbonates of active metals (like Na₂CO₃) often react with halogens in a disproportionation reaction, acting as a base to neutralize the acid produced, resulting in the evolution of CO₂.