Number of Sodium Bromide Molecules in Balanced Reaction of Bromine with Sodium Carbonate

To determine the number of sodium bromide ($\text{NaBr}$) molecules involved in the balanced chemical equation, we can use the redox balancing method (oxidation number method).

Step 1: Identify the oxidation states

In the given disproportionation reaction, elemental bromine ($\text{Br}_2$) is simultaneously oxidized and reduced:

• Unreacted Bromine ($\text{Br}_2$): Oxidation state of $\text{Br} = 0$
• Sodium Bromide ($\text{NaBr}$): Oxidation state of $\text{Br} = -1$ (reduction)
• Sodium Bromate ($\text{NaBrO}_3$): Oxidation state of $\text{Br} = +5$ (oxidation)

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Step 2: Balance the oxidation and reduction half-reactions

1. Oxidation half-reaction:
   $\text{Br}_2^0 \longrightarrow 2\text{Br}^{+5} + 10e^-$

2. Reduction half-reaction:
   $\text{Br}_2^0 + 2e^- \longrightarrow 2\text{Br}^{-1}$

To balance the number of electrons lost and gained, multiply the reduction half-reaction by $5$:
$5\text{Br}_2 + 10e^- \longrightarrow 10\text{Br}^-$

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Step 3: Combine the half-reactions

Adding the two balanced half-reactions gives:
$\text{Br}_2 + 5\text{Br}_2 \longrightarrow 10\text{Br}^- + 2\text{BrO}_3^-$
$6\text{Br}_2 \longrightarrow 10\text{Br}^- + 2\text{BrO}_3^-$

Dividing the entire ionic equation by $2$ to express it in the simplest whole-number ratio:
$3\text{Br}_2 \longrightarrow 5\text{Br}^- + \text{BrO}_3^-$

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Step 4: Balance the cations and anions to get the molecular equation

• To balance the negative charges on the right-hand side ($5\text{Br}^-$ and $1\text{BrO}_3^-$), we need $6\text{Na}^+$ ions.
• These $6\text{Na}^+$ ions are provided by $3\text{Na}_2\text{CO}_3$ on the reactant side.
• The carbonate ions ($\text{CO}_3^{2-}$) decompose to produce $\text{CO}_2$ gas and oxygen atoms required for $\text{NaBrO}_3$:

$3\text{Br}_2 + 3\text{Na}_2\text{CO}_3 \longrightarrow 5\text{NaBr} + \text{NaBrO}_3 + 3\text{CO}_2$

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Step 5: Verify mass and charge balance

• Bromine ($\text{Br}$): $3 \times 2 = 6$ on reactants side; $5 (1) + 1 (1) = 6$ on products side.
• Sodium ($\text{Na}$): $3 \times 2 = 6$ on reactants side; $5 (1) + 1 (1) = 6$ on products side.
• Carbon ($\text{C}$): $3$ on reactants side; $3$ on products side.
• Oxygen ($\text{O}$): $3 \times 3 = 9$ on reactants side; $3 (\text{in } \text{NaBrO}_3) + 3 \times 2 (\text{in } \text{CO}_2) = 9$ on products side.

The equation is fully balanced. The coefficient of sodium bromide ($\text{NaBr}$) in the balanced equation is $5$.