Question

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

$\mathrm{Lead}\mathrm{Dioxide}+(\mathrm{conc}.)\mathrm{hydrochloric}\mathrm{acid}\to \mathrm{lead}\mathrm{chloride}+\mathrm{water}+\mathrm{chlorine}$

Answer 1

The given word equation is $\mathrm{Lead}\mathrm{Dioxide}+(\mathrm{conc}.)\mathrm{hydrochloric}\mathrm{acid}\to \mathrm{lead}\mathrm{chloride}+\mathrm{water}+\mathrm{chlorine}$

The symbol equation for the given word equation is ${\mathrm{PbO}}_2+\mathrm{HCl}\to {\mathrm{PbCl}}_2+{\mathrm{H}}_2\mathrm{O}+{\mathrm{Cl}}_2$.

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	No. of atoms of reactant(L.H.S)	No. of atoms of products(R.H.S)
$\mathrm{Pb}$	1	1
$\mathrm{O}$	2	1
$\mathrm{H}$	1	2
$\mathrm{Cl}$	1	2+2

Thus, the number of 'Pb' atoms are equal on both sides. At the same time, the number of ‘O’ atoms , 'H' atoms and ‘Cl’ atoms are not equal.

On inspection, To balance the atoms, put coefficient 4 before $\mathrm{HCl}$ on the reactant side. and 2 before ${\mathrm{H}}_2\mathrm{O}$on product side

The partially balanced equation may be written as

${\mathrm{PbO}}_2+4.\mathrm{HCl}\to {\mathrm{PbCl}}_2+2.{\mathrm{H}}_2\mathrm{O}+{\mathrm{Cl}}_2$

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

$\underset{\mathrm{Lead}\mathrm{dioxide}}{{\mathrm{PbO}}_2\left(\mathrm{s}\right)}+\underset{\mathrm{Hydrochloric}\mathrm{acod}}{4.\mathrm{HCl}(\mathrm{aq}.)}\to \underset{\mathrm{Lead}\mathrm{chloride}}{{\mathrm{PbCl}}_2(\mathrm{aq}.)}+\underset{\mathrm{Water}}{2.{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}+\underset{\mathrm{Chlorine}}{{\mathrm{Cl}}_2\left(\mathrm{g}\right)}$