Question

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

Answer 1

Hint The number of atoms on the right hand side and left hand side must be equal. Multiply the atoms with a natural number to get a balanced equation.

Complete step by step solution:
So in the question, an unbalanced chemical equation of a chemical reaction is given. And we are asked to balance the given equation. First before going into the solution part we should know what a chemical equation is, how it is written and what data it gives us.
A chemical equation is simply the representation of the reaction happening in the reaction vessel. From the time we are studying about different elements we are very much familiar that each element has its own chemical symbol or notation for representing the species. The chemical notation may be the first letter or two letters taken from the name of the elements. The first letter of the element is always written in capital form.

Now we should know the components in an equation. The equation mainly has two parts which is, the left hand side called as the reactant side where the chemical notations of the reactants in the chemical equation is represented. And in the right hand side the products formed in the chemical reaction are represented. There is an arrow placed in between the reactant side and product side, the reaction conditions are written above the arrow.

Now we have to know what information does a balanced chemical equation give. A balanced chemical equation means an equation in which both the reactant and product side have equal number of atoms and the equation gives us an idea about number of moles of substance reacting and number of moles of different products formed.

Now let’s move to the solution part. Count the number of atoms on each side of the equation.
$ Al+{{H}_{2}}S{{O}_{4}}\to A{{l}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}+{{H}_{2}}$
LHS: Al -1, H-2, S-1, O-4
RHS: Al-2, H-2, S-3, O-12

Now let’s balance the equation, first let's start with Al .Multiply 2 with Al on the reactant side, so that the number of Al atoms will be the same on both sides. And the equation becomes,
$ 2Al+{{H}_{2}}S{{O}_{4}}\to A{{l}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}+{{H}_{2}}$
Now let’s balance the S atom, for that multiply $ {{H}_{2}}S{{O}_{4}}$ with 3 and the equation becomes,
$ 2Al+3{{H}_{2}}S{{O}_{4}}\to A{{l}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}+{{H}_{2}}$
So almost all atoms except H are balanced on both the sides. To balance the H atom in the product side 4 more H atoms are required and hence multiply $ {{H}_{2}}$ with 3 which gives a total of 6 H in both the sides.
And the final balanced equation will be,
$ 2Al+3{{H}_{2}}S{{O}_{4}}\to A{{l}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}+3{{H}_{2}}$

Note: We should remember that while balancing a chemical equation only the coefficients attached to the atom should be changed, the subscripts must not be changed as it changes the stoichiometry of the compound.