Question

How would you balance the following equation ${S_8} + {O_2} \to S{O_3}$ ?

Answer 1

In a reaction, the number of each element should be the same on reactant and product sides. Balancing a chemical reaction involves equating the numbers of reactants and products on each side.

Complete step by step answer: The given reaction is,
${S_8} + {O_2} \to S{O_3}$
First let us check whether the equation is not balanced or not. There are eight sulphur atoms in the reactant side. But on the product side, there is only one sulphur atom. Hence the number of sulphur atoms is not balanced in the given equation. Also the number of oxygen atoms on the reactant side is two while the number of oxygen atoms on the product side is three. Hence the number of oxygen atoms is also not balanced. So the equation is not balanced.
Now let us balance the equation. To make the number of sulphur atoms equal on both sides, add the number eight near the compound $S{O_3}$. Then we get the following equation.
${S_8} + {O_2} \to 8S{O_3}$
The number of oxygen on the right side became $24$ . But on the left side only two oxygen atoms are present. Hence let us add a $12$ on the oxygen present in the left side of the equation. The equation thus becomes,
${S_8} + 12{O_2} \to 8S{O_3}$
Now the number of oxygen and sulphur atoms present on both sides are equal. Hence the equation is balanced.

Note:
Balancing a chemical reaction is very important in chemistry. It helps to determine the stoichiometry of compounds involved in a reaction. In industries, in order to determine the limiting reagent, balanced chemical equations are necessary.