Question

How would you balance ${{\text{S}}_{\text{8}}} + {{\text{F}}_{\text{2}}} \to {\text{S}}{{\text{F}}_{\text{6}}}$?

Answer 1

Hint: To balance the given reaction we must know that in a balanced chemical equation the number of all the atoms of the elements present in the chemical reaction are equal. A symbolic representation of a chemical reaction which has an equal number of atoms of each element in the product and the reactant is known as a balanced chemical equation.

Complete step by step solution:
We know that a symbolic representation of a chemical reaction which has an equal number of atoms of each element in the product and the reactant is known as a balanced chemical equation.
To balance the given reaction we must know that in a balanced chemical equation the number of all the atoms of the elements present in the chemical reaction are equal.
The given reaction is as follows:
${{\text{S}}_{\text{8}}} + {{\text{F}}_{\text{2}}} \to {\text{S}}{{\text{F}}_{\text{6}}}$
In the given reaction, there are 8 sulphur atoms on the reactant side and only 1 sulphur atom on the product side.
Thus, to balance the number of sulphur atoms change the coefficient of the product i.e. ${\text{S}}{{\text{F}}_{\text{6}}}$ to 8. Thus, the reaction is as follows:
${{\text{S}}_{\text{8}}} + {{\text{F}}_{\text{2}}} \to 8{\text{S}}{{\text{F}}_{\text{6}}}$
Now, there are 2 fluorine atoms on the reactant side and 48 fluorine atoms on the product side.
Thus, to balance the number of fluorine atoms change the coefficient of the reactant i.e. ${{\text{F}}_{\text{2}}}$ to 24. Thus, the reaction is as follows:
${{\text{S}}_{\text{8}}} + 24{{\text{F}}_{\text{2}}} \to 8{\text{S}}{{\text{F}}_{\text{6}}}$
Thus, the balanced chemical equation is as follows:
${{\text{S}}_{\text{8}}} + 24{{\text{F}}_{\text{2}}} \to 8{\text{S}}{{\text{F}}_{\text{6}}}$

Note: Remember that to balance any chemical equation first count each atom on the reactant and product side. Then balance each atom by changing the coefficients. Balanced chemical equations are important in determining the stoichiometry of the reaction.