Question

Balance the following chemical equations including the physical state.
- ${{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{6}}}{\text{ }} \to {\text{ }}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH + C}}{{\text{O}}_{\text{2}}}$
- ${\text{Fe + }}{{\text{O}}_{\text{2}}} \to {\text{ F}}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}$
- ${\text{N}}{{\text{H}}_{\text{3}}}{\text{ + C}}{{\text{l}}_{\text{2}}} \to {\text{ }}{{\text{N}}_{\text{2}}}{{\text{H}}_{\text{4}}}{\text{ + N}}{{\text{H}}_{\text{4}}}{\text{Cl}}$
- ${\text{Na + }}{{\text{H}}_{\text{2}}}{\text{O }} \to {\text{ NaOH + }}{{\text{H}}_{\text{2}}}$

Answer 1

In the above question, we are asked to balance the chemical equations and add their physical state. Here, we have to compare the no. of atoms present on both the side of the reaction in order to get the balance equation and also write the general physical state they are present in.

Complete Step by step answer
Since, in the above question we have to balance all the four equations, let us start it from the beginning.
In equation a, we have:
${{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{6}}}{\text{ }} \to {\text{ }}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH + C}}{{\text{O}}_{\text{2}}}$
Let us first balance the H atom, 12 atoms are present in reactant side any only 6 are present on the product side so, multiplying 2 as a coefficient of ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH}}$ we get:
${{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{6}}}{\text{ }} \to {\text{ 2}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH + C}}{{\text{O}}_{\text{2}}}$
Now we have 6 oxygen in the reactant side and 4 on product side, so by multiplying 2 as coefficient of ${\text{C}}{{\text{O}}_{\text{2}}}$ we get:
${{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{6}}}{\text{ }} \to {\text{ 2}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH + 2C}}{{\text{O}}_{\text{2}}}$
Now, we can check that number of carbon atoms is also same on both the side, so the balanced equation is:
${{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{6}}}{\text{ (s)}} \to {\text{ 2}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH (l) + 2C}}{{\text{O}}_{\text{2}}}{\text{(g)}}$
In equation b, we have:
${\text{Fe + }}{{\text{O}}_{\text{2}}} \to {\text{ F}}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}$
Let us first balance the oxygen atom, we have 2 oxygen atoms on the reactant side and 3 oxygen on the product side. Hence, we have to balance it by multiply 2 before ${\text{F}}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ and 3 before ${{\text{O}}_{\text{2}}}$ . Hence we the equation as:
${\text{Fe + 3}}{{\text{O}}_{\text{2}}} \to {\text{ 2F}}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}$
Now, we have 1 Fe atom on reactant side but 4 on product side, so we have to multiply 4 before ${\text{Fe}}$ , hence we get:
${\text{4Fe + 3}}{{\text{O}}_{\text{2}}} \to {\text{ 2F}}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}$
So, balance chemical equation is:
${\text{4Fe (s) + 3}}{{\text{O}}_{\text{2}}}{\text{(g) }} \to {\text{ 2F}}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}{\text{(s)}}$
In equation c, we have:
${\text{N}}{{\text{H}}_{\text{3}}}{\text{ + C}}{{\text{l}}_{\text{2}}} \to {\text{ }}{{\text{N}}_{\text{2}}}{{\text{H}}_{\text{4}}}{\text{ + N}}{{\text{H}}_{\text{4}}}{\text{Cl}}$
Let us balance Cl atom by multiplying 2 to ${\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}$ , hence we have :
${\text{N}}{{\text{H}}_{\text{3}}}{\text{ + C}}{{\text{l}}_{\text{2}}} \to {\text{ }}{{\text{N}}_{\text{2}}}{{\text{H}}_{\text{4}}}{\text{ + 2N}}{{\text{H}}_{\text{4}}}{\text{Cl}}$
There are 1 N on reactant side but 4 on product side, so we will add 4 as coefficient of ${\text{N}}{{\text{H}}_{\text{3}}}$ ,hence we get:
${\text{4N}}{{\text{H}}_{\text{3}}}{\text{ + C}}{{\text{l}}_{\text{2}}} \to {\text{ }}{{\text{N}}_{\text{2}}}{{\text{H}}_{\text{4}}}{\text{ + 2N}}{{\text{H}}_{\text{4}}}{\text{Cl}}$
We can see that H atom is already balanced, so the balanced equation is:
${\text{4N}}{{\text{H}}_{\text{3}}}{\text{(g) + C}}{{\text{l}}_{\text{2}}}{\text{(g)}} \to {\text{ }}{{\text{N}}_{\text{2}}}{{\text{H}}_{\text{4}}}{\text{(l) + 2N}}{{\text{H}}_{\text{4}}}{\text{Cl(s)}}$
In equation d, we have:
${\text{Na + }}{{\text{H}}_{\text{2}}}{\text{O }} \to {\text{ NaOH + }}{{\text{H}}_{\text{2}}}$
Since, 2 H atom are present on reactant and 3 on product side, we have to multiply 2 before ${{\text{H}}_{\text{2}}}{\text{O}}$ and ${\text{NaOH }}$ , hence we get:
${\text{Na + 2}}{{\text{H}}_{\text{2}}}{\text{O }} \to {\text{ 2NaOH + }}{{\text{H}}_{\text{2}}}$
Since, only Na atom is not balanced, we can add 2 before Na, hence, the balanced chemical equation:
${\text{2Na(s) + 2}}{{\text{H}}_{\text{2}}}{\text{O(l) }} \to {\text{ 2NaOH(s) + }}{{\text{H}}_{\text{2}}}{\text{(g)}}$
Hence, we have balanced all the equations.

Note
In these types of questions where we have multiple reactants and products, we should first try to balance the atom which is present only in one of the reactants and products and then we can proceed further.
We should balance an equation in order to satisfy the law of conservation of mass which states that mass can neither be created nor be destroyed.
If an equation is not balanced, then no information can be gathered from the equation. An equation which is not balanced is also known as a skeletal equation.