Question

How do I balance this equation?
\[{\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}} + \,{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\, \to \,\,{\text{A}}{{\text{l}}_{\text{2}}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{3}}} + \,{{\text{H}}_{\text{2}}}{\text{O}}\]

Answer 1

Hint:Here, the reaction is given between the aluminium oxide and sulphuric acid which gives aluminium sulfate and water.When a chemical reaction is written using the chemical formulae of the compound or species involved in the reaction then it is known as a chemical equation.When there are the same numbers of atoms of elements on reactant as well as the product side of the chemical equation then it is a balanced chemical equation.

Complete step by step answer:
The stepwise balancing of the given reaction is as follows:
To balance the chemical equation the first step is to write the chemical equation.
\[{\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}} + \,{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\, \to \,\,{\text{A}}{{\text{l}}_{\text{2}}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{3}}} + \,{{\text{H}}_{\text{2}}}{\text{O}}\]
Now, count each type of atom present on both the sides of the reaction.
\[{\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}} + \,\,\,\,\,{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,{\text{A}}{{\text{l}}_{\text{2}}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{3}}} + \,{{\text{H}}_{\text{2}}}{\text{O}}\,\]
\[{\text{2Al}}\,\,\,\,{\text{2H}}\,\,\,\,{\text{1S}}\,\,\,7{\text{O}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{2Al}}\,\,\,\,{\text{3S 13O}}\,\,\,\,{\text{2H}}\,\]
Now, aluminium and sulphur are present in only one species on reactant as well as on the product side. Therefore, balance its number first.
The numbers of aluminium on reactant as well as on product side are same hence, no need to balanced it.
Now, balance sulphur as there is one sulphur on the reactant side but 3 on the product side. Hence, multiply \[{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\] by 3.
\[{\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}} + \,\,\,\,\,3{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,{\text{A}}{{\text{l}}_{\text{2}}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{3}}} + \,{{\text{H}}_{\text{2}}}{\text{O}}\,\]
\[{\text{2Al}}\,\,\,\,6{\text{H}}\,\,\,\,3{\text{S}}\,\,\,15{\text{O}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{2Al}}\,\,\,\,{\text{3S 13O}}\,\,\,\,{\text{2H}}\,\]
Here, sulphur atoms are balanced.
Now, balance the hydrogen atoms as there is 6 hydrogen on the reactant side but 2 on the product side. Hence, multiply \[{{\text{H}}_{\text{2}}}{\text{O}}\] on the product side by 3.
\[{\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}} + \,\,\,\,\,3{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,{\text{A}}{{\text{l}}_{\text{2}}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{3}}} + \,3{{\text{H}}_{\text{2}}}{\text{O}}\]
\[{\text{2Al}}\,\,\,\,6{\text{H}}\,\,\,\,3{\text{S}}\,\,\,15{\text{O}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{2Al}}\,\,\,\,{\text{3S 15O}}\,\,\,\,6{\text{H}}\,\,\]
Thus, hydrogen atoms are balanced.
Here, we can see that after balancing hydrogen atoms, oxygen atoms also get balanced.
Now, if we count all the atoms in the equation their number is the same on both the sides.
Hence, the chemical equation gets balanced.
\[{\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}} + \,\,3{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\,\, \to \,{\text{A}}{{\text{l}}_{\text{2}}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{3}}} + \,3{{\text{H}}_{\text{2}}}{\text{O}}\]
This is the balanced chemical equation.

Note:The balanced chemical equation is very important to determine the stoichiometry of the reaction. The stoichiometric ratio obtained between the reactants and products is important to determine the yield of the species formed in the reaction.
The stoichiometric ratio is nothing but the mole ratio between the reactants and the products of the reaction.