Question

How would you balance \[N{{a}_{2}}S{{O}_{3}}+{{S}_{8}}\to N{{a}_{2}}{{S}_{2}}{{O}_{3}}\] ?

Answer 1

Hint: A chemical equation is symbolic representation of chemical reaction from which reactant as well as the product are given by their respective chemical formula. The given example of a chemical equation is \[N{{a}_{2}}S{{O}_{3}}+{{S}_{8}}\to N{{a}_{2}}{{S}_{2}}{{O}_{3}}\]. The reactant side is part of the chemical equation in order to the left of the $'\to '$ symbol whereas the product side is part to the right of the arrow symbol.

Complete step-by-step answer:
The steps for balancing a chemical reaction are the first step that must be followed while balancing the chemical equation to obtain a complete un-balanced chemical equation. The reaction between Sodium sulfite and Octasulfur.

Step 1: The unbalanced equation needs to be obtained from the chemical formula of reactant and product. The chemical formula of Sodium sulfite \[\left( N{{a}_{2}}S{{O}_{3}} \right)\] reacts with Octasulfur $\left( {{S}_{8}} \right)$ to form Sodium thiosulfate \[\left( N{{a}_{2}}{{S}_{2}}{{O}_{3}} \right)\] and the unbalanced chemical equation can be written as: \[N{{a}_{2}}S{{O}_{3}}+{{S}_{8}}\to N{{a}_{2}}{{S}_{2}}{{O}_{3}}\]

Step 2: The total numbers of each element of atoms on reactant side and product side needs to be compared. For this the number of atoms on each side can be given as follows.
Unbalanced Chemical Reaction: \[N{{a}_{2}}S{{O}_{3}}+{{S}_{8}}\to N{{a}_{2}}{{S}_{2}}{{O}_{3}}\]

Reactant Side	Product side
$2$ sodium atom from \[N{{a}_{2}}S{{O}_{3}}\]	$2$ sodium atom from \[N{{a}_{2}}{{S}_{2}}{{O}_{3}}\]
$9$ sulphur atoms; $1$ from \[N{{a}_{2}}S{{O}_{3}}\] and $8$ from \[{{S}_{8}}\]	$2$ Sulphur atoms from \[N{{a}_{2}}{{S}_{2}}{{O}_{3}}\]
$3$ oxygen atoms from \[N{{a}_{2}}S{{O}_{3}}\]	$3$ Oxygen atoms from \[N{{a}_{2}}{{S}_{2}}{{O}_{3}}\]

Step 3: Now that we have stoichiometric coefficients which are added to molecules and are containing elements which have different numbers of atoms in reactant side as well as product side Here all coefficients must balance the number of atoms by each side.

Now number of atom of the element on reactant and product side needs t be updated it's important to note numbers of atoms of element with one species which needs to be obtained by multiplying stoichiometric coefficient with total number of atom of that element present in single molecule of that species.

That’s is by multiplying by \[8N{{a}_{2}}{{S}_{2}}{{O}_{3}}\] in reactant and in product by \[N{{a}_{2}}{{S}_{2}}{{O}_{3}}\]


Unbalanced Chemical Reaction: \[8N{{a}_{2}}S{{O}_{3}}+{{S}_{8}}\to 8N{{a}_{2}}{{S}_{2}}{{O}_{3}}\]

Reactant Side	Product side
$16$ sodium atom from \[8N{{a}_{2}}S{{O}_{3}}\]	$16$ sodium atom from \[8N{{a}_{2}}{{S}_{2}}{{O}_{3}}\]
$16$ sulphur atoms; $8$ from \[8N{{a}_{2}}S{{O}_{3}}\] and $8$ from ${{S}_{8}}$	$16$ sulphur atoms from \[8N{{a}_{2}}{{S}_{2}}{{O}_{3}}\]
$24$ oxygen atoms from \[8N{{a}_{2}}S{{O}_{3}}\]	$24$ oxygen atoms from \[8N{{a}_{2}}{{S}_{2}}{{O}_{3}}\]

Step 4: Once all of our compounds from reactants and products are individual elements are balanced and the total number of atoms of each element on reactants and products side is compared again and if there are no inequalities therefore we can say that the chemical reaction is a balanced chemical equation. In the above given reaction every element now has an equal number of atoms in reactants and products side.

Therefore, Balanced Chemical Equation is \[8N{{a}_{2}}S{{O}_{3}}+{{S}_{8}}\to 8N{{a}_{2}}{{S}_{2}}{{O}_{3}}\]

Note: The things which we have to remember when balancing chemical reactions are: The same element in equal numbers i.e. for the reaction to be balanced it needs to have the same number of each element on each other side of equation. The mass of the reactant has to equal the mass of the product and we have to remember the law of conservation of mass.