Question

How would you balance $Al\text{ + }{{\text{H}}_{2}}S{{O}_{4}}\text{ = A}{{\text{l}}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}\text{ + }{{\text{H}}_{2}}$?

Answer 1

Hint: To balance a chemical equation, make sure that the number of atoms of each element on the reactant side is the same as the number of atoms of each element on the product side. Atoms on both sides should be equal. If they are not balanced, multiply the less element with less number of atoms by a suitable coefficient.

Complete answer:
When aluminium reacts with Sulphuric acid it gives Aluminium sulphide and hydrogen gas is liberated.
The given reaction is:
$Al\text{ + }{{\text{H}}_{2}}S{{O}_{4}}\text{ }\to \text{ A}{{\text{l}}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}\text{ + }{{\text{H}}_{2}}$
To balance the atoms of each element, first, let’s see the number of atoms of each element.

Element/molecule	No of atoms on the Reactant side	No of atoms on the product side
Al	1	2
H	2	2
$S{{O}_{4}}$	1	3

Now we can see that the number of hydrogen atoms is balanced on both sides. But, the number of Aluminium and Sulphate atoms is unbalanced.
To equate the number of atoms on the reactant we will multiply the aluminium by 2 and $S{{O}_{4}}$by 3. $S{{O}_{4}}$is a part of ${{\text{H}}_{2}}S{{O}_{4}}$ so 3 will be placed before ${{\text{H}}_{2}}S{{O}_{4}}$
The resultant equation will be:
$2Al\text{ + 3}{{\text{H}}_{2}}S{{O}_{4}}\text{ }\to \text{ A}{{\text{l}}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}\text{ + }{{\text{H}}_{2}}$
But as we multiplied ${{\text{H}}_{2}}S{{O}_{4}}$by 3, the number of atoms of H on the reactant side changed to 6 whereas the number of atoms of H on the product side is still 2. So to balance the number of atoms of H on the product side we’ll multiply H on the right side by 3.
So, the equation will be:
$2Al\text{ + 3}{{\text{H}}_{2}}S{{O}_{4}}\text{ }\to \text{ A}{{\text{l}}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}\text{ + 3}{{\text{H}}_{2}}$
The balanced chemical equation is $2Al\text{ + 3}{{\text{H}}_{2}}S{{O}_{4}}\text{ }\to \text{ A}{{\text{l}}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}\text{ + 3}{{\text{H}}_{2}}$.

Note:
While balancing one element in a molecule, the number of atoms of the other elements tends to change. So always check the number of atoms of all elements after adding any coefficient. To find a suitable coefficient for balancing the equation, find the LCM of the number of atoms on both sides of the equation.