Question

Write the balanced equation for the following chemical reactions:
(i) $hydrogen+chlorine\to hydrogen\text{ }chloride$
(ii) $barium\text{ }chloride+aluminum\text{ }sulphate\to barium\text{ }sulphate+aluminum\text{ }chloride$
(iii) $sodium+water\to sodium\text{ }hydroxide+hydrogen$

Answer 1

Balancing of the reaction means that the number of moles of each of the atoms involved in that very chemical reaction should have the equal no of moles on both sides i.e. the reactants and the product sides. Now balance the equations.

Complete answer:
First, we should know what a chemical reaction is. The reaction in which the reactants react with each other to form two or more products is known as the chemical reaction. The reactants are written on the left hand side and the products are written on the right-hand side and there is an arrow from left to the right symbolizing a complete chemical reaction. Example: consider the general reaction as:
$\text{A+B}\to \text{C+D}$
Here, A and B are the reactants and C and D are the products and the arrow indicates the direction in which the chemical reaction occurs.
In this, the chemical equation is said to be balanced, when it has the equal no of atoms of each element on both the sides i.e. on the reactant and the products side.
Now, we will balance the given chemical reactions with the equal no of moles on both the reactant and the product side.
(a) $hydrogen+chlorine\to hydrogen\text{ }chloride$
${{\text{H}}_{2}}\text{+ C}{{\text{l}}_{2}}\to HCl\text{ }$ ---------(1)

The no of moles of each atom on the reactant side	The no of moles of each atom on the product side
H= 1Cl=1	HCl=1

Since, there are 2 moles of ${{H}_{2}}$and $C{{l}_{2}}$ on the reactant side , so we will add 2 in front of HCl on the product side, in eq (1), we get;
Balanced reaction:${{\text{H}}_{2}}\text{+ C}{{\text{l}}_{2}}\to 2HCl\text{ }$
Similarly, we can balance the rest of the reactions too.
(ii) $barium\text{ }chloride+aluminum\text{ }sulphate\to barium\text{ }sulphate+aluminum\text{ }chloride$
$BaC{{l}_{2}}\text{+ A}{{\text{l}}_{2}}{{\text{(S}{{\text{O}}_{4}}\text{)}}_{3}}\to \text{ BaS}{{\text{O}}_{4}}\text{+AlC}{{\text{l}}_{3}}$------(2)
Now, consider the reaction (2)

The no of moles of each atom on the reactant side	The no of moles of each atom on the product side
Ba= 1	Ba= 1
Cl=2	Cl=3
Al=2	Al=1
S=3	S=1
O=12	O=4

Since, there are 2 Aluminum atoms and 3sulphur atoms on the reactants side then, we will add 3 in front of the $BaS{{O}_{4}}$ and add 2 in front of $AlC{{l}_{3}}$ to balance the equation. Then the equation (2) becomes:
$BaC{{l}_{2}}\text{+ A}{{\text{l}}_{2}}{{\text{(S}{{\text{O}}_{4}}\text{)}}_{3}}\to \text{ 3BaS}{{\text{O}}_{4}}\text{+2AlC}{{\text{l}}_{3}}$------(2 a)

The no of moles of each atom on the reactant side	The no of moles of each atom on the product side
Ba= 1	Ba= 3
Cl=2	Cl=6
Al=2	Al=2
S=3	S=3
O=12	O=12

Now, consider the reaction (2 a)
Since, there are 3 Barium atoms and 6 chlorine atoms on the products side then, we will add 3 in front of the $BaC{{l}_{2}}$ to balance the equation. Then the equation (2 a) becomes:
$3BaC{{l}_{2}}\text{+ A}{{\text{l}}_{2}}{{\text{(S}{{\text{O}}_{4}}\text{)}}_{3}}\to \text{ 3BaS}{{\text{O}}_{4}}\text{+2AlC}{{\text{l}}_{3}}$---------(2 b)
Hence, the equation (2b) is the fully balanced equation.
Similarly, we can balance the reaction (c).
iii) $sodium+water\to sodium\text{ }hydroxide+hydrogen$
$\text{Na + }{{\text{H}}_{2}}\text{O}\to \text{ NaOH+}{{\text{H}}_{2}}$
Balanced reaction: $\text{2Na + 2}{{\text{H}}_{2}}\text{O}\to \text{ 2NaOH+}{{\text{H}}_{2}}$

Note:
After balancing, check whether the no of moles of each atom are same on both the sides or not and balancing the chemical reaction is very important because it helps to conserve the mass of that very atom/substance or the compound.