Question

Balance the following equation:
${\text{C}}{{\text{l}}_{\text{2}}}{\text{ + NaOH}} \to {\text{NaCl + NaCl}}{{\text{O}}_{\text{3}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}$

Answer 1

Hint For balancing any chemical equation, we have to maintain equal molecularity of each atom present on the reactant side as well as on the product side of the given chemical reaction.

Complete step by step solution:
In the question following chemical equation is given:
${\text{C}}{{\text{l}}_{\text{2}}}{\text{ + NaOH}} \to {\text{NaCl + NaCl}}{{\text{O}}_{\text{3}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}$
For balancing the above given equation we have to consider following points in mind:
-First we calculate the number of each atom on the left hand side as well as on the right hand side of the given chemical equation.
-So on the left hand side or reactant side and on the right hand side or product side of the chemical reaction two chlorine ( ${\text{Cl}}$ ) atoms are present, which is correct.
-On the left hand side one atom of sodium ( ${\text{Na}}$ ) is present & on the right hand side two sodium atoms are present which is not correct.
-On the left hand side one atom of oxygen ( ${\text{O}}$ ) is present & on the right hand side four oxygen atoms are present which is not correct.
- Similarly, on the left side only one hydrogen atom ( ${\text{H}}$ ) and on the right side two hydrogen atoms ( ${\text{H}}$ ) are present which is not correct.
So, we will balance the above given reaction by the following manner so that each atom will have the same molecularity on the reactant side as well as on the product of the reaction. And balanced equation is shown as:
${\text{3C}}{{\text{l}}_{\text{2}}}{\text{ + 6NaOH}} \to 5{\text{NaCl + NaCl}}{{\text{O}}_{\text{3}}}{\text{ + 3}}{{\text{H}}_{\text{2}}}{\text{O}}$
-In the above balanced equation six chlorine atoms ( ${\text{Cl}}$ ), six sodium atoms ( ${\text{Na}}$ ), six oxygen atoms ( ${\text{O}}$ ) and six hydrogen ( ${\text{H}}$ ) atoms are present on the left hand side as well as on the right hand side of the chemical reaction also.

Note: Here some of you may think order of reaction and molecularity of reaction are the same things but that will be true only for the elementary (single step reaction) reactions only.