Question

Write the chemical formula and hence balance the following word equation:

$\mathrm{Aluminium}\mathrm{oxide}+(\mathrm{dil}.)\mathrm{sulphuric}\mathrm{acid}\to \mathrm{Aluminium}\mathrm{Sulphate}+\mathrm{water}$

Answer 1

The Given word equation is$\mathrm{Aluminium}\mathrm{oxide}+(\mathrm{dil}.)\mathrm{sulphuric}\mathrm{acid}\to \mathrm{Aluminium}\mathrm{Sulphate}+\mathrm{water}$.

The symbol equation of the given word equation is ${\mathrm{Al}}_2{\mathrm{O}}_3+{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3+{\mathrm{H}}_2\mathrm{O}$

The balancing of the equations is done in the following steps :
Let us count the number of atoms of all the elements in the reactants and products on both sides of the equation.

Element/ion	No. of atoms of reactant(L.H.S)	No. of atoms of products(R.H.S)
$\mathrm{Al}$	2	2
$\mathrm{O}$	3	1
${\mathrm{SO}}_4$	1	3
$\mathrm{H}$	2	2

Thus, the number of ‘Al' atoms and ’H ‘are equal on both sides. At the same time, the number of ’O' and ${\mathrm{SO}}_4$ ions are not equal.

On inspection, To balance the atoms, put coefficient 3 before ${\mathrm{H}}_2{\mathrm{SO}}_4$ on the reactant side. And coefficient 3 before ${\mathrm{H}}_2\mathrm{O}$ on the product side.

The partially balanced equation may be written as

${\mathrm{Al}}_2{\mathrm{O}}_3+3{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3+3{\mathrm{H}}_2\mathrm{O}$

On final inspection, the number of atoms of all the elements on both sides of the equation is equal. Therefore, the equation is balanced.

$\underset{\mathrm{Aluminum}\mathrm{oxide}}{{\mathrm{Al}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)}+\underset{\mathrm{Sulfuric}\mathrm{Acid}}{3{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)}\to \underset{\mathrm{Aluminum}\mathrm{Sulfate}}{{\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3\left(\mathrm{aq}\right)}+\underset{\mathrm{Water}}{3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$