Question

Write the balanced equation for the following chemical reaction: $\text{Barium chloride + Aluminium sulphate }\xrightarrow{{}}\text{ Barium sulphate + Aluminium chloride}$
A. $\text{3BaC}{{\text{l}}_{\text{2}}}\left( \text{aq} \right)\text{ + A}{{\text{l}}_{\text{2}}}{{\left( \text{S}{{\text{O}}_{\text{4}}} \right)}_{\text{3}}}\left( \text{aq} \right)\text{ }\xrightarrow{{}}\text{ 3BaS}{{\text{O}}_{\text{4}}}\left( \text{s} \right)\text{ + 2A1C}{{\text{l}}_{\text{3}}}\left( \text{aq} \right)$
B. $\text{BaC}{{\text{l}}_{\text{2}}}\left( \text{aq} \right)\text{+ A}{{\text{l}}_{\text{2}}}{{\left( \text{S}{{\text{O}}_{\text{4}}} \right)}_{\text{3}}}\left( \text{aq} \right)\text{ }\xrightarrow{{}}\text{ BaS}{{\text{O}}_{\text{4}}}\left( \text{s} \right)\text{ + AlC}{{\text{l}}_{\text{3}}}\left( \text{aq} \right)$
C. $\text{3BaC}{{\text{l}}_{\text{2}}}\left( \text{aq} \right)\text{+ A}{{\text{l}}_{\text{2}}}{{\left( \text{S}{{\text{O}}_{\text{4}}} \right)}_{\text{3}}}\left( \text{aq} \right)\text{ }\xrightarrow{{}}\text{ BaS}{{\text{O}}_{\text{4}}}\left( \text{s} \right)\text{ +2A1C}{{\text{l}}_{\text{3}}}\left( \text{aq} \right)\text{ }\\!\\!~\\!\\!\text{ }$
D. $\text{BaC}{{\text{l}}_{\text{2}}}\left( \text{aq} \right)\text{ + A}{{\text{l}}_{\text{2}}}{{\left( \text{S}{{\text{O}}_{\text{4}}} \right)}_{\text{3}}}\left( \text{aq} \right)\text{ }\xrightarrow{{}}\text{ 3BaS}{{\text{O}}_{\text{4}}}\left( \text{s} \right)\text{ + 2A1C}{{\text{l}}_{\text{3}}}\left( \text{aq} \right)$

Answer 1

To balance a chemical equation, we need to know the meaning of it at first. By balancing chemical equations, we mean making sure that the number of the different atoms of elements in the reactants side is equal to the side of the products.

Step by step answer:
To begin the balancing of the chemical equation, we need to follow certain steps. Here they are listed one by one:
First, we need to write formulas for the reactants and the products that are provided in the question. The products and the reactants that are given is:
$\text{Barium chloride + Aluminium sulphate }\xrightarrow{{}}\text{ Barium sulphate + Aluminium chloride}$
Now, changing the compounds into their chemical formulas:
$\text{BaC}{{\text{l}}_{\text{2}}}\text{ + A}{{\text{l}}_{\text{2}}}{{\text{(S}{{\text{O}}_{\text{4}}}\text{)}}_{\text{3}}}\xrightarrow{{}}\text{ BaS}{{\text{O}}_{\text{4}}}\text{ + AlC}{{\text{l}}_{\text{3}}}$
Secondly, we are required to count the number of atoms of each element present on the reactant and the product side, so that we can equalise them.
As we can see that the number of atoms of each element is not equal on both sides (that is reactant and product) we need to multiply them with numbers which can help us in making the atoms on both sides of the chemical reaction equal.
So, now it is time for trial and error. Chemical equation balancing is a process which needs to be on the basis of hit and trial. So, unless we find the correct number to multiple the compounds, we need to go on with the process.
In option A, we can see that both the reactants and the products consist of elements which all have equal number of atoms, on both the sides.
$\text{3BaC}{{\text{l}}_{\text{2}}}\left( \text{aq} \right)\text{ + A}{{\text{l}}_{\text{2}}}{{\left( \text{S}{{\text{O}}_{\text{4}}} \right)}_{\text{3}}}\left( \text{aq} \right)\text{ }\xrightarrow{{}}\text{ 3BaS}{{\text{O}}_{\text{4}}}\left( \text{s} \right)\text{ + 2A1C}{{\text{l}}_{\text{3}}}\left( \text{aq} \right)$
In option B, the reaction that is given is:
$\text{BaC}{{\text{l}}_{\text{2}}}\left( \text{aq} \right)\text{+ A}{{\text{l}}_{\text{2}}}{{\left( \text{S}{{\text{O}}_{\text{4}}} \right)}_{\text{3}}}\left( \text{aq} \right)\text{ }\xrightarrow{{}}\text{ BaS}{{\text{O}}_{\text{4}}}\left( \text{s} \right)\text{ + AlC}{{\text{l}}_{\text{3}}}\left( \text{aq} \right)$
From the above reaction we can see that the number of Al atoms is not similar on the reactants and products side. Even O atoms are not similar. So, this is not a balanced equation.
In option C, we see the reaction:
$\text{3BaC}{{\text{l}}_{\text{2}}}\left( \text{aq} \right)\text{+ A}{{\text{l}}_{\text{2}}}{{\left( \text{S}{{\text{O}}_{\text{4}}} \right)}_{\text{3}}}\left( \text{aq} \right)\text{ }\xrightarrow{{}}\text{ BaS}{{\text{O}}_{\text{4}}}\left( \text{s} \right)\text{ +2A1C}{{\text{l}}_{\text{3}}}\left( \text{aq} \right)\text{ }\\!\\!~\\!\\!\text{ }$
In this reaction, we see that Ba is not balanced. Moreover, S is also not balanced. There are 3 atoms of Ba on the reactants side and 1 atom on the products side. And in case of S, there are 3 atoms on the reactants side, and1 atom on the products side. So, this reaction is not balanced.
In the last option D, we have
$\text{BaC}{{\text{l}}_{\text{2}}}\left( \text{aq} \right)\text{ + A}{{\text{l}}_{\text{2}}}{{\left( \text{S}{{\text{O}}_{\text{4}}} \right)}_{\text{3}}}\left( \text{aq} \right)\text{ }\xrightarrow{{}}\text{ 3BaS}{{\text{O}}_{\text{4}}}\left( \text{s} \right)\text{ + 2A1C}{{\text{l}}_{\text{3}}}\left( \text{aq} \right)$
In this reaction we have Ba unbalanced, and even O. There are 3 atoms of Ba in the products side, but 1 atom on the reactants side. With O, there are 12 atoms on the reactants side, and 4 atoms on the products side.
Therefore, the correct option is Option A.
Note: The main aim of chemical equation balancing is to preserve the law of conservation of mass. The law of conservation of law states that matter can neither be created nor can be destroyed, in chemical reactions. So, the number of atoms of each element, on both the sides of the reactions will be the same.