Question

How do you write a balanced chemical equation for the fermentation of sucrose $({{C}_{12}}{{H}_{22}}{{O}_{11}})$ by yeast in which the aqueous sugar reacts with water to form aqueous sugar reacts with water to form aqueous ethyl alcohol $({{C}_{2}}{{H}_{5}}OH)$ and carbon dioxide gas?

Answer 1

Hint: Firstly we have to figure out the chemical equation of above given chemical reaction. In a balanced chemical equation the number of particular elements in a reactant should be equal to the number of elements in the product.

Complete step by step answer:
- Chemical equation is the representation of a chemical reaction where the reactants react to give products. In a chemical equation the substance that is left to the arrow is called reactant and the substance that is to the right of the arrow is called product. The law of conservation of mass attributes for balancing a chemical equation. The law of conservation of mass which states that the atoms are neither created nor destroyed. Applying the law of conservation of mass to a chemical, there is no atom destroyed from the reactant side nor created in the product, so the number of elements in the reactant is equal to the number of elements in the product. In a balanced chemical equation the number of particular elements in a reactant should be equal to the number of elements in the product.
- The fermentation of sucrose by yeasts in which the aqueous sugar reacts with water to form aqueous ethyl alcohol and carbon dioxide gas. The chemical equation of this reaction is:
\[{{C}_{12}}{{H}_{22}}{{O}_{11}}+{{H}_{2}}O\to {{C}_{2}}{{H}_{5}}OH+C{{O}_{2}}\]
This chemical equation is unbalanced.
- Let's try a trial and error method, where we will check if the number of elements in the reactant is equal to the number of elements in the product and balance the equation.
- The number of C atoms in the reactant side is 12 but the number of C atoms in the product side is 3. In order to balance it lets add 4 before ${{C}_{2}}{{H}_{5}}OH$and $C{{O}_{2}}$.
\[{{C}_{12}}{{H}_{22}}{{O}_{11}}+{{H}_{2}}O\to 4{{C}_{2}}{{H}_{5}}OH+4C{{O}_{2}}\]
-Now, all the atoms are equal in number on both sides.
Thus, the balanced chemical equation of given reaction is:
\[{{C}_{12}}{{H}_{22}}{{O}_{11}}+{{H}_{2}}O\to 4{{C}_{2}}{{H}_{5}}OH+4C{{O}_{2}}\]

Note: When we are taking a number of elements, if the element has both subscript and coefficient in order to get a number of elements we should multiply subscript and coefficient. Only one simple principle is that the number of a particular element in the reactant should be equal to the number of elements in the product.