Question

Aluminium react with sulphuric acid $\left( {{H_2}S{O_4}} \right)$ to produce aluminium sulphate $A{l_2}{\left( {S{O_4}} \right)_3}$ and hydrogen $\left( {{H_2}} \right)$. Complete the balance chemical equation for this reaction?

Answer 1

Balanced chemical equation is used to express a chemical equation in which an equal number of particular atoms is present in reactant as well as in product. Balancing of chemical equations is based on various approaches like hit and trial method, partial equation method, ion electron method and oxidation number method.

Complete answer:
There are some basic rules which needed to be followed while balancing the chemical reaction –
$\to$ Firstly, we have to write the complete chemical reaction including all of its reactants and products.
$\to$ Write the chemical reaction in which all the elements are represented by their chemical formula and symbol.
$\to$ Start balancing the equation with the chemical formula which contains the highest number of atoms.
$\to$Calculate the number of different atoms present in reaction at both the sides of reaction.
$\to$ Now calculate the net difference between the atom in reactant side and product side.
$\to$ Multiply with suitable integers before the chemical formula to balance the atoms on both sides.
$\to$ At last verify the chemical reaction by calculating the number of atoms on both sides of the chemical reaction.
Chemical reaction of aluminium with sulphuric acid to form aluminium sulphate and hydrogen is expressed as-
$Al + {H_2}S{O_4} \to A{l_2}{\left( {S{O_4}} \right)_3} + {H_2}$

REACTANT SIDE	PRODUCT SIDE
ELEMENT	NUMBER
$Al$	$1$
$H$	$2$
$S$	$1$
$O$	$4$

To balance the atoms of aluminium we have to multiply by $2$ on the reactant side.
To balance the atoms of sulphur we have to multiply by $3$ on the reactant side.
To balance the atoms of oxygen we have to multiply by $3$ on the reactant side.
As we see, atoms of hydrogen are already balanced in the chemical reaction.
Hence, the final configuration of the chemical reaction will become-

REACTANT SIDE	PRODUCT SIDE
NUMBER	ELEMENT
$1 \times 2 = 2$	$Al$
$2$	$H$
$1 \times 3 = 3$	$S$
$4 \times 3 = 12$	$O$

From the above table we see that when we multiply aluminium with $2$ and sulphuric acid with $3$ our chemical reaction becomes balanced.
$\Rightarrow$ So, the final balanced chemical equation become-
$2Al + 3{H_2}S{O_4} \to A{l_2}{\left( {S{O_4}} \right)_3} + {H_2}$

Note:
During the balancing of chemical reactions if reactant is present in its gaseous state, then change it into atomic state. Aluminium metal readily reacts with acids to undergo the process of displacement reaction to form salt of aluminium and release hydrogen gas.