Question

How do I balance this equation?
${{C}_{4}}{{H}_{8}}S+{{O}_{2}}\to C{{O}_{2}}+{{H}_{2}}O+S{{O}_{3}}$

Answer 1

A reaction in which the number of atoms involved in a reaction is equal for each element on both the reactant side and the product side is known as a balanced chemical equation. A chemical reaction should be balanced to satisfy the law of conservation of matter.

Complete step-by-step answer: In a chemical equation, there are two types of numbers that determine the number of atoms and molecules. These are coefficients and subscripts.
Coefficients indicate the number of molecules of a substance that is used or produced or produced in a chemical reaction.
Subscripts are a part of the chemical formulas that shows the ratio of atoms of a molecule and determines its chemical identity. They are not changed while balancing an equation.
A chemical equation can be balanced by using these simple steps. Let us take an example of combustion of heptane or ${{C}_{7}}{{H}_{16}}$.
- First, we write the unbalanced equation of the reaction.
${{C}_{4}}{{H}_{8}}S+{{O}_{2}}\to C{{O}_{2}}+{{H}_{2}}O+S{{O}_{3}}$
- Now we compare the number of atoms of each element on the product and the reactant side.
Here, there are 4 carbon atoms, 8 hydrogen atoms and 1 sulfur atom from ${{C}_{4}}{{H}_{8}}S$, and 2 oxygen atoms from ${{O}_{2}}$ on the reactant side. There are 1 carbon atom and 2 oxygen atoms from $C{{O}_{2}}$ 2 hydrogen atoms and one oxygen atom from ${{H}_{2}}O$, and 1 sulfur atom and 3 oxygen atoms from $S{{O}_{3}}$ on the product side.
- Now we add stoichiometric coefficient to the molecules containing elements that have a different number of atoms on the reactant and product side. Oxygen and hydrogen are balanced last.
The most complex substance is balanced first. Here ${{C}_{4}}{{H}_{8}}S$ is the most complex structure. So, there must be 4 atoms of carbon and 1 atom of sulfur on both the reactant and product side.
${{C}_{4}}{{H}_{8}}S+{{O}_{2}}\to 4C{{O}_{2}}+{{H}_{2}}O+S{{O}_{3}}$
- Now we balance the coefficients of hydrogen. Since there are 8 hydrogen atoms from ${{C}_{4}}{{H}_{8}}S$, there must be 8 hydrogen atoms from ${{H}_{2}}O$.
${{C}_{4}}{{H}_{8}}S+{{O}_{2}}\to 4C{{O}_{2}}+4{{H}_{2}}O+S{{O}_{3}}$
- Balance the polyatomic ions, if present, as a unit. This reaction has no polyatomic ions.
- Then we balance the remaining atoms. There are 15 oxygen atoms on the product side as compared to 2 oxygen atoms from reactants. So, this can be balanced by
${{C}_{4}}{{H}_{8}}S+\dfrac{15}{2}{{O}_{2}}\to 4C{{O}_{2}}+4{{H}_{2}}O+S{{O}_{3}}$
- Now, this equation can be simplified by multiplying the whole equation by 2. So, we get
$2{{C}_{4}}{{H}_{8}}S+15{{O}_{2}}\to 8C{{O}_{2}}+8{{H}_{2}}O+2S{{O}_{3}}$
- The equation is now balanced, as the number of atoms of each molecule are equal on both the product side as well as the reactant side.

Note: Redox reactions can be balanced by two methods. Oxidation number method, which involves using the difference in the oxidizing agent oxidation number and the reducing agent oxidation number, and the half-reaction method, which is based on the division of the redox reactions into the reduction half and the oxidation half.