Question

Reaction of $\mathrm{Br}_{2}$ with $\mathrm{Na}_{2} \mathrm{CO}_{3}$ in aqueous solution gives sodium bromide and sodium bromate with evolution of $\mathrm{CO}_{2}$ gas. The number of sodium bromide molecules involved in the balanced chemical equation is:
(A)5
(B)4
(C)3
(D)2

Answer 1

Hint: The balanced chemical reaction is in accordance with the law of conservation of mass. According to which, the mass can neither be created nor destroyed in a chemical reaction. But the total mass of the atoms of the molecules on the reactant side is always equal to the mass of atoms of the molecules on the product side. The balancing is done by introducing the coefficients in front of the molecules.

Complete step by step solution:
-The chemical equation is defined as the representation of the chemical reaction in terms of substance. These substances can be reactant or the product. The chemical reactions obey the law of conservation of mass. According to the law, the mass cannot be created nor be destroyed, but it can be changed from one composition to the other.
-The balanced chemical reactions are the application of the law of conservation of mass. In the balanced chemical equation, the number of atoms or the number of moles of an atom of all the reactants and the product is equal on both sides of the reaction.
-In other words, the number of moles of an atom on the reactant side always equals the number of moles of the same atom on the product side.
-Now we have asked for balance as a chemical reaction.
-Reaction of $\mathrm{Br}_{2}$ with $\mathrm{Na}_{2} \mathrm{CO}_{3}$ in aqueous solution gives sodium bromide and sodium bromate with evolution of $\mathrm{CO}_{2}$ gas.
$$
\mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{Br}_{2} \rightarrow \mathrm{NaBr}+\mathrm{NaBrO}_{3}+\mathrm{CO}_{2}
$$
-Now let us balance this reaction.
-We have 3 oxygen atoms in the reactant side. But we have 5 oxygen atoms in the product side. Thus, let us add 3 in front of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ and add 3 as coefficient in front of $\mathrm{CO}_{2}$. Thus, the oxygens are balanced.
-We have 6 Na atoms in the reactant. So let us add the coefficient 5 in front of NaBr. Thus, the Na atoms are balanced.

-We have $2 \mathrm{Br}$ atoms in the reactant side and $6 \mathrm{Br}$ atoms in the product side. Thus we add the coefficient 3 in the $\mathrm{Br}_{2}$ in the reactant side. Thus the $\mathrm{Br}$ atoms are balanced.
The Balanced chemical reaction is:
$$
3 \mathrm{Na}_{2} \mathrm{CO}_{3}+3 \mathrm{Br}_{2} \rightarrow 5 \mathrm{NaBr}+\mathrm{NaBrO}_{3}+3 \mathrm{CO}_{2}
$$

The number of sodium bromide molecules involved in the balanced chemical equation is 5. Hence, the correct option is (A).

Note: We never try to balance the equation by changing the subscript of the molecule. The subscript is unique and has characteristics of a specific molecule, which decides the number of atoms in molecules, the stability, reactivity, etc. Changing the subscript would result in the new molecules which may lead to the undesired product. Thus, always balance the reaction by considering the coefficient in front of atoms. In this way, the equation is balanced correctly.