Question

Aluminium hydroxide reacts with sulphuric acid to form aluminium sulphate and water. Write a balanced equation for this reaction.

Answer 1

Hint: Keep in mind that aluminum hydroxide is a compound that is basic in nature which contains aluminum. On the other hand, sulphuric acid's nature is quite corrosive and it has a high acidic composition. It functions as an oxidizing and dehydrating agent at higher concentrations. So, in this case, one compound is an acid and the other is a base compound.

Complete answer:
Let us understand the reactants and products formed in the equation first, then we shall proceed to form its balanced equation.
It is given that the reactants of this reaction are:
Aluminium hydroxide, Sulphuric acid
We are also given the products:
Aluminium sulphate, Water
The reaction involves reactants where one is acidic (Sulphuric acid) and the other is basic (Aluminium hydroxide), as a result of which the products formed were a salt (Aluminium sulphate) and water.
Such a reaction can be termed as a Neutralization reaction, due to the involvement of an acid and a base.
The given reaction can be written in terms of their chemical formulae:
 $ Al{(OH)_3}{\;_{(aq)}} + {H_2}S{O_4}{\;_{(aq)}} \to A{l_2}{(S{O_4})_3}{\;_{(aq)}} + {H_2}O{\;_{(l)}} $
Here the compounds are:
 $ Al{(OH)_3} $ - Aluminium hydroxide
 $ {H_2}S{O_4} $ - Sulphuric Acid
 $ A{l_2}{(S{O_4})_3} $ - Aluminium Sulphate
 $ {H_2}O $ - Water
In the reaction we see that the products were formed by:
- Aluminium took the place of hydrogen in sulphuric acid to form aluminium sulphate
- The hydrogen from sulphuric acid combined with the hydroxide present in aluminium hydroxide to form water.
So there have been two displacements that took place in the above reaction, to form the products hence it is a double displacement reaction.
Now to balance this equation, we must make sure that the number of atoms of each element is the same on the reactant side and the product side.
We will balance the hydrogen and oxygen atoms only at the end since they will keep varying.
So first we shall try to balance the metal aluminium: $ Al $
On the product side, there are two atoms of aluminium so we can put two atoms of aluminium on the reactant side also:
 $ 2Al{(OH)_3}{\;_{(aq)}} + {H_2}S{O_4}{\;_{(aq)}} \to A{l_2}{(S{O_4})_3}{\;_{(aq)}} + {H_2}O{\;_{(l)}} $
Now we can try to balance the element sulphur: $ S $
There are three sulphur atoms on the product side so we can put three sulphur atoms on the reactant side also:
 $ 2Al{(OH)_3}{\;_{(aq)}} + 3{H_2}S{O_4}{\;_{(aq)}} \to A{l_2}{(S{O_4})_3}{\;_{(aq)}} + {H_2}O{\;_{(l)}} $
Now we see that the reactant side has twelve atoms of hydrogen and eighteen atoms of oxygen. Simultaneously on the product side we have two atoms of hydrogen and thirteen atoms of oxygen.
We will try to balance hydrogen first since it is easier:
 $ 2Al{(OH)_3}{\;_{(aq)}} + 3{H_2}S{O_4}{\;_{(aq)}} \to A{l_2}{(S{O_4})_3}{\;_{(aq)}} + 6{H_2}O{\;_{(l)}} $
Now that we have balanced hydrogen, we can move on to balancing oxygen.
On the reactant side there are eighteen oxygen atoms and now on the product side we have eighteen atoms. So we see that oxygen is also balanced in the equation.
Therefore the final balanced chemical equation for the given reaction is:
 $ 2Al{(OH)_3}{\;_{(aq)}} + 3{H_2}S{O_4}{\;_{(aq)}} \to A{l_2}{(S{O_4})_3}{\;_{(aq)}} + 6{H_2}O{\;_{(l)}} $ .

Note:
In the given scenario we discussed double displacement reactions so now we will see about single displacement reactions. When any element from the first compound replaces another element of the second compound, a single displacement reaction occurs. A metal can only replace another metal, similarly a nonmetal can only replace another nonmetal.