Question

Balance the following unbalanced reaction:

$\mathrm{PbS}+{\mathrm{H}}_2{\mathrm{O}}_2\to {\mathrm{PbSO}}_4+{\mathrm{H}}_2\mathrm{O}$

Answer 1

Steps involved in balancing a chemical reaction:

Step 1: Write unbalanced equation.

$\mathrm{PbS}+{\mathrm{H}}_2{\mathrm{O}}_2\to {\mathrm{PbSO}}_4+{\mathrm{H}}_2\mathrm{O}$

Step 2: Write down the number of atoms of each elements present on both reactant as well as product side.

Element Reactant side Product side

$\mathrm{Pb}$ $1$ $1$

$\mathrm{S}$ $1$ $1$

$\mathrm{H}$ $2$ $2$

$\mathrm{O}$ $2$ $5$

Step 3: Add coefficients to balance the number of atoms at both the sides

Firstly balance $\mathrm{O}$ at reactant side by adding $4$ before ${\mathrm{H}}_2{\mathrm{O}}_2$.

$\mathrm{PbS}+4{\mathrm{H}}_2{\mathrm{O}}_2\to {\mathrm{PbSO}}_4+{\mathrm{H}}_2\mathrm{O}$

Secondly balance $\mathrm{O}$ and $\mathrm{H}$ at product side by adding $4$ before ${\mathrm{H}}_2\mathrm{O}$.

$\mathrm{PbS}+4{\mathrm{H}}_2{\mathrm{O}}_2\to {\mathrm{PbSO}}_4+4{\mathrm{H}}_2\mathrm{O}$

Balanced equation is:

$\mathrm{PbS}+4{\mathrm{H}}_2{\mathrm{O}}_2\to {\mathrm{PbSO}}_4+4{\mathrm{H}}_2\mathrm{O}$

Lead(II)sulphide Hydrogen peroxide Lead(II)sulphate Water