Question

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

($\mathrm{Copper}+(\mathrm{conc}.)\mathrm{Sulfuric}\mathrm{acid}\to \mathrm{Copper}\mathrm{sulphate}+\mathrm{Sulphur}\mathrm{dioxide}+\mathrm{Water}$

Answer 1

The given word equation is $\mathrm{Copper}+(\mathrm{conc}.)\mathrm{Sulfuric}\mathrm{acid}\to \mathrm{Copper}\mathrm{sulphate}+\mathrm{Sulphur}\mathrm{dioxide}+\mathrm{Water}$

The symbol equation for the given word equation is $\mathrm{Cu}+{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{CuSO}}_4+{\mathrm{SO}}_2+{\mathrm{H}}_2\mathrm{O}$

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)
$\mathrm{Cu}$	1	1
$\mathrm{O}$	4	4+2+1
$\mathrm{S}$	2	1+1
$\mathrm{H}$	2	2

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

On inspection, To balance the atoms, put coefficient 2 before ${\mathrm{H}}_2{\mathrm{SO}}_4$ on the reactant side. And coefficient 2 before ${\mathrm{H}}_2\mathrm{O}$ on the product side.

The partially balanced equation may be written as

$\mathrm{Cu}+2{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{CuSO}}_4+{\mathrm{SO}}_2+2{\mathrm{H}}_2\mathrm{O}$

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

$\underset{\mathrm{Copper}}{\mathrm{Cu}\left(\mathrm{s}\right)}+\underset{\mathrm{Sulfuric}\mathrm{acid}}{2{\mathrm{H}}_2{\mathrm{SO}}_4(\mathrm{aq}.)}\to \underset{\mathrm{Copper}\mathrm{Sulfate}}{{\mathrm{CuSO}}_4(\mathrm{aq}.)}+\underset{\mathrm{Sulfur}\mathrm{dioxide}}{{\mathrm{SO}}_2\left(\mathrm{g}\right)}+\underset{\mathrm{Water}}{2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$