Question

How do you write a balanced equation for the combustion of liquid methyl alcohol $C{{H}_{3}}OH$?

Answer 1

Hint: First of all, write the chemical reaction in terms of chemical formulas. Then write down the no. of atoms at reactant side and no. of atoms at product side and further try to make the numbers equal which will result in a balanced equation.

Complete step-by-step answer:A balanced chemical equation is the usage of chemical symbols to show the molecules and atoms in a chemical reaction. The reactants are represented on the left hand side of the equation and the products are on the right hand side. Coefficients give information about the number of molecules involved and the subscripts give us information about the number of atoms present in each molecule.
Let us write down the chemical equation as follows:-
-$C{{H}_{3}}OH+{{O}_{2}}\xrightarrow{{}}C{{O}_{2}}+{{H}_{2}}O$
Now let us write down the no. of atoms on each side of the reaction:-
Reactant side: - Product side:-
C= 1 C= 1
H= 4 H= 2
O= 3 O= 3
As we can see that, no. of C and O atoms are same on both side but no. of H atom is different. Therefore, to equalize the no. of H atom, we will multiply ${{H}_{2}}O$ with 2.
-$C{{H}_{3}}OH+{{O}_{2}}\xrightarrow{{}}C{{O}_{2}}+2{{H}_{2}}O$
Now again let us write down the no. of atoms on each side of the reaction:-
Reactant side: - Product side:-
C= 1 C= 1
H= 4 H= 4
O= 3 O= 4
As we can see that, no. of C and H atoms are same on both side but no. of O atom is different. Therefore, to equalize the no. of O atom, we will multiply ${{O}_{2}}$ with 3/2 (We multiplied ${{O}_{2}}$ with 3/2 because we can change its no. of atom easily without affecting other atoms at the end.)
-$C{{H}_{3}}OH+\dfrac{3}{2}{{O}_{2}}\xrightarrow{{}}C{{O}_{2}}+2{{H}_{2}}O$
Now again let us write down the no. of atoms on each side of the reaction:-
Reactant side: - Product side:-
C= 1 C= 1
H= 4 H= 4
O= 4 O= 4
Since all the atoms on both the sides are same, therefore $C{{H}_{3}}OH+\dfrac{3}{2}{{O}_{2}}\xrightarrow{{}}C{{O}_{2}}+2{{H}_{2}}O$ is the balanced chemical equation.

Note:Whenever we multiply molecules with some number, always write down the no. of atoms on both sides because we multiply the number with not only a particular atom but with all the atoms that molecule is composed of. Also we can multiply this $C{{H}_{3}}OH+\dfrac{3}{2}{{O}_{2}}\xrightarrow{{}}C{{O}_{2}}+2{{H}_{2}}O$ by 2 so as to get all coefficients as integers which are very helpful in solutions.
$2C{{H}_{3}}OH+3{{O}_{2}}\xrightarrow{{}}2C{{O}_{2}}+4{{H}_{2}}O$