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
580.5k+ views
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.
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.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

