Question

How do you balance $HCl+Ba{{\left( OH \right)}_{2}}\to BaC{{l}_{2}}+{{H}_{2}}O$?

Answer 1

Hint The balanced equation will have an equal number of atoms on the both sides of the arrow. Multiply the unbalanced equation with an appropriate number which is called as the coefficient of the atoms.

Complete step by step solution:
In the question, it is asked how we will balance the given chemical equation. Before going into the balancing equation, let’s first discuss the chemical equation.
A chemical reaction is represented in the form of a chemical reaction. The chemical elements which take part in the chemical reaction are written in the notation form along with the subscripts according to the number of atoms that are involved in the chemical combination.
The chemical equation consists of the right hand side and left hand side. The right hand side of the chemical equation is called the reactant side, where the reactants that are involved in the chemical reaction are represented. The left hand side of the chemical reaction represents the product formed in the chemical reaction and is called the product side.
Let’s study the steps involved in balancing the chemical equation by solving the above equation given.
The above given equation is a chemical equation for the displacement reaction.
To balance the chemical equation we should first sort out the number of atoms involved in the chemical reaction along with their respective numbers on both sides.
The equation is: $HCl+Ba{{\left( OH \right)}_{2}}\to BaC{{l}_{2}}+{{H}_{2}}O$
Now write the number of each atom present in the reactant side.
LHS:
$\text{H=3}$
$\text{Cl=1}$
$\text{Ba=1}$
$\text{O=2}$
For the polyatomic species which is represented in the parentheses, if there is any subscript after the parentheses, then the subscript is applicable for all the atoms present inside the parentheses.
Now let’s write the number of different atoms present in the product side.
RHS:
$\text{H=2}$
$\text{Cl=2}$
$\text{Ba=1}$
$\text{O=1}$
Now let’s start balancing the equation, first let’s balance the Cl atom by multiplying it with 2.
This changes the equation as,
$2HCl+Ba{{\left( OH \right)}_{2}}\to BaC{{l}_{2}}+{{H}_{2}}O$
And now the number of atoms has changed in the equation, at this point there are four H atoms ($\text{H=4}$) in the reactant side and only 2 in the product side.
And now let’s balance the number of H on both the sides by multiplying the ${{\text{H}}_{\text{2}}}\text{O}$with 2.
On giving the coefficient as 2 for ${{\text{H}}_{\text{2}}}\text{O}$, the equation becomes,
$2HCl+Ba{{\left( OH \right)}_{2}}\to BaC{{l}_{2}}+2{{H}_{2}}O$
Now check the number of atoms on both the sides of the arrow.
The various atoms present in the equation are thus balanced as in the equation, the number of each atom present in the chemical equation is equal on the reactant and product side.

Note: Always be careful while dealing with the polyatomic ion species as we make mistakes while counting the all the atoms in the parentheses and the subscript present is applicable for all the atoms present in the polyatomic species. We should only change the coefficients of the atoms and not the subscripts of the atoms to balance the equation. If the subscripts are changed it changes the stoichiometry of the compounds.