Question

Write the balanced equation for the following chemical reactions-
(a) $\text{Hydrogen + Chlorine }\to \text{ Hydrogen chloride}$
(b) $\text{Barium chloride + Aluminium sulphate }\to \text{ Barium sulphate + Aluminium chloride}$
(c) $\text{Sodium + water }\to \text{ Sodium hydroxide + Hydrogen}$

Answer 1

Hint: Balanced chemical equation is a chemical equation in which the number of each type of atom is equal on the two sides of the equation. The states of the chemical species should also be mentioned.

Complete answer:
First write the chemical equation by identifying the reactants, products and their chemical formulae.
-The coefficients prior to the symbols of chemical species indicate the number of moles of a substance produced or used in the chemical reaction.
-Balanced chemical equation is defined as a chemical equation in which the number of each type of atom is equal on the two sides of the equation.
-To balance a chemical equation, we first determine the number of atoms of each element on both sides of the equation. The equation is then balanced by adjusting the coefficients.
-The state symbol of each substance is also mentioned in a chemical reaction. The symbols (s), (l) and (g) represent the solid, liquid and gaseous states respectively and the symbol (aq) represents an aqueous solution.
(a) $\text{Hydrogen + Chlorine }\to \text{ Hydrogen chloride}$
Symbol of hydrogen molecule: ${{H}_{2}}$
Symbol of chlorine molecule: $C{{l}_{2}}$
Symbol of hydrogen chloride: HCl
Therefore the chemical equation is-
\[{{H}_{2}}(g)+C{{l}_{2}}(g)\to HCl(g)\]

Element	No. of atoms on Reactant side	No. of atoms on Product side
H	2	1
Cl	2	1

We can easily write the coefficient 2 before HCl on the product side to make the equation balanced such as-
\[{{H}_{2}}(g)+C{{l}_{2}}(g)\to 2HCl(g)\]

Element	No. of atoms on Reactant side	No. of atoms on Product side
H	2	2
Cl	2	2

As the number of atoms of H and Cl is the same on the reactant and product side, the equation is balanced. Also, all three entities are gaseous in nature.
(b) $\text{Barium chloride + Aluminium sulphate }\to \text{ Barium sulphate + Aluminium chloride}$
Symbol of barium chloride: $BaC{{l}_{2}}$
Symbol of aluminium sulphate: \[A{{l}_{2}}{{(S{{O}_{4}})}_{3}}\]
Symbol of barium sulphate: $BaS{{O}_{4}}$
Symbol of aluminium chloride: $AlC{{l}_{3}}$
Therefore the chemical equation is-
\[BaC{{l}_{2}}(aq)+A{{l}_{2}}{{(S{{O}_{4}})}_{3}}(aq)\to BaS{{O}_{4}}(s)+AlC{{l}_{3}}(aq)\]

Element	No. of atoms on Reactant side	No. of atoms on Product side
Ba	1	1
Cl	2	3
Al	2	1
$S{{O}_{4}}$	3	1

We get $BaS{{O}_{4}}$ as a precipitate in this reaction therefore we write solid state, rest all are in aqueous state.
We can balance the number of Cl atoms by making the total number of Cl atoms as 6. So use the coefficient 3 before $BaC{{l}_{2}}$ and 2 before $AlC{{l}_{3}}$.with this step, the Al atoms also get balanced.
\[3BaC{{l}_{2}}(aq)+A{{l}_{2}}{{(S{{O}_{4}})}_{3}}(aq)\to BaS{{O}_{4}}(s)+2AlC{{l}_{3}}(aq)\]

Element	No. of atoms on Reactant side	No. of atoms on Product side
Ba	3	1
Cl	6	6
Al	2	2
$S{{O}_{4}}$	3	1

Now to balance number of Ba and $S{{O}_{4}}$ atoms, we can add the coefficient 3 before $BaS{{O}_{4}}$
Now the equation becomes-
\[3BaC{{l}_{2}}(aq)+A{{l}_{2}}{{(S{{O}_{4}})}_{3}}(aq)\to 3BaS{{O}_{4}}(s)+2AlC{{l}_{3}}(aq)\]

Element	No. of atoms on Reactant side	No. of atoms on Product side
Ba	3	3
Cl	6	6
Al	2	2
$S{{O}_{4}}$	3	3

All atoms are now balanced.
(c) $\text{Sodium + water }\to \text{ Sodium hydroxide + Hydrogen}$
Symbol of sodium: Na
Symbol of water : ${{H}_{2}}O$
Symbol of sodium hydroxide : NaOH
Symbol of hydrogen molecule: ${{H}_{2}}$
Therefore the chemical equation is-\[\]
\[Na(s)+{{H}_{2}}O(l)\to NaOH(aq)+{{H}_{2}}(g)\]

Element	No. of atoms on Reactant side	No. of atoms on Product side
Na	1	1
H	2	3
O	1	1

To balance the number of H atoms, we can add the coefficient 2 before ${{H}_{2}}O$ and 2 before NaOH to make the total number of H atoms as 4. We get the equation as-
\[Na(s)+2{{H}_{2}}O(l)\to 2NaOH(aq)+{{H}_{2}}(g)\]

Element	No. of atoms on Reactant side	No. of atoms on Product side
Na	1	2
H	4	4
O	2	2

Now to balance Na atoms, we can use 2 before Na on the reactant side. We get-
\[2Na(s)+2{{H}_{2}}O(l)\to 2NaOH(aq)+{{H}_{2}}(g)\]

Element	No. of atoms on Reactant side	No. of atoms on Product side
Na	2	2
H	4	4
O	2	2

Now the equation is completely balanced.

Additional information:
The symbol $\downarrow $ is used to indicate the formation of a precipitate.
The symbol $\uparrow $ is used to show the release of a gas.

Note:
Please note that coefficient is a whole number that appears in front of a formula in a balanced chemical equation while Subscript is part of the chemical formulas of the reactants and products that refers to the number of atoms of the preceding element.