Question

How would you balance the following chemical equation?
$\text{CaS}{{\text{O}}_{4}}+\text{AlC}{{\text{l}}_{3}}\to \text{CaC}{{\text{l}}_{2}}+\text{A}{{\text{l}}_{2}}{{\left( \text{S}{{\text{O}}_{4}} \right)}_{3}}$

Answer 1

: Major changes occurring during a chemical reaction are represented in the form of chemical equations with the help of formulas of different species involved. A chemical equation is said to be balanced when the number of atoms of each species involved on the left-hand side of the equation is equal to that on the right-hand side.

Complete step-by-step answer: According to the Law of conservation of Mass, in an isolated system the matter cannot be created nor be destroyed. That is why, during a chemical reaction also, the number of species reacting with each other must be equal to the number of species that are getting produced.
Now, let us see what steps we need to follow to balance a chemical equation.

1. – First look at the given chemical equation and try to find out which elements are not balanced.
   $\text{CaS}{{\text{O}}_{4}}+\text{AlC}{{\text{l}}_{3}}\to \text{CaC}{{\text{l}}_{2}}+\text{A}{{\text{l}}_{2}}{{\left( \text{S}{{\text{O}}_{4}} \right)}_{3}}$
   In the given equation, we have a sulphate ion, so we will treat it as a single entity. We can see here that there is an unequal number of atoms of sulphate, aluminium, and chlorine on both sides of the equation.

2. – Then we should start balancing sulphate by multiplying it by 3 on the reactant side.
   $\text{3CaS}{{\text{O}}_{4}}+\text{AlC}{{\text{l}}_{3}}\to \text{CaC}{{\text{l}}_{2}}+\text{A}{{\text{l}}_{2}}{{\left( \text{S}{{\text{O}}_{4}} \right)}_{3}}$

3. – Now, the sulphate is balanced. But there are 3 calcium atoms on the reactant side while only 1 on the product side. So, there must be 3 atoms of calcium on the left-hand side too. So, let's multiply $\text{CaC}{{\text{l}}_{2}}$ by 3.
   $\text{3CaS}{{\text{O}}_{4}}+\text{AlC}{{\text{l}}_{3}}\to 3\text{CaC}{{\text{l}}_{2}}+\text{A}{{\text{l}}_{2}}{{\left( \text{S}{{\text{O}}_{4}} \right)}_{3}}$

4. – Now let’s note down the number of atoms on both sides of the equation.

5. – There are 2 atoms of aluminium on the product side whereas there is only 1 atom on the reactant side. So, if we multiply $\text{AlC}{{\text{l}}_{3}}$ by 2 on the right-hand side of the equation, then our equation becomes:
   $\text{3CaS}{{\text{O}}_{4}}+2\text{AlC}{{\text{l}}_{3}}\to 3\text{CaC}{{\text{l}}_{2}}+\text{A}{{\text{l}}_{2}}{{\left( \text{S}{{\text{O}}_{4}} \right)}_{3}}$

6. – Now we can see that along with aluminium, chlorine also gets balanced. Let's again check the total number of atoms on both sides.

7. – The number of atoms of each species is now equal on both sides of the equation. Our equation is now balanced.
   Hence, by this trial and error method, we can balance the chemical equation.

Note: The coefficient is the whole number by which we multiply the element or compound to balance the chemical equation and the subscript is the number that comes in the formula of a compound which cannot be altered. While calculating the total number of atoms, we should multiply its coefficient with its subscript number to get the exact value.