Question

Lead(II) nitrate on heating with an aqueous solution of potassium iodide, gives a yellow precipitate. Write a balanced equation for the reaction and name the type of reaction.

Answer 1

1. Type of reaction

Combination reaction: Reaction in which two or more reactants combine to form a single product. $\mathrm{A}+\mathrm{B}\to \mathrm{C}$

Decomposition reaction: Reaction in which reactant is broken down into two or more products in the presence of heat, light or electricity. $\mathrm{A}\to \mathrm{B}+\mathrm{C}$

Displacement reaction: Reaction in which one of the higher reactive element displaces less reactive element from its solution. $\mathrm{A}+\mathrm{BC}\to \mathrm{AC}+\mathrm{B}$ ( Here A is more reactive than B)

Double displacement reaction: Reaction in which ions of both the reactants interchange their place and form new compounds. $\mathrm{AB}+\mathrm{CD}\to \mathrm{AD}+\mathrm{CB}$

2. Reaction:

$\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2+\mathrm{KI}\stackrel{\mathrm{heat}}{\to }{\mathrm{PbI}}_2+{\mathrm{KNO}}_3$

Lead(II) nitrate Potassium iodide Lead iodide Potassium nitrate

(Yellow)

On heating lead(II) nitrate with potassium iodide double displacement reaction takes place both $\mathrm{Pb}$ and $\mathrm{K}$ interchange their position.

3. Balancing steps:

Balancing steps:
Step 1: Write the unbalanced equation.
$\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2+\mathrm{KI}\stackrel{\mathrm{heat}}{\to }{\mathrm{PbI}}_2+{\mathrm{KNO}}_3$

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
$\mathrm{Pb}$ $1$ $1$
$\mathrm{N}$ $2$ $1$
$\mathrm{O}$ $6$ $3$

$\mathrm{K}$ $1$ $1$

$\mathrm{I}$ $1$ $1$

Step 3: Add coefficients to balance the number of atoms on both sides.
i. Balance $\mathrm{N}$ and $\mathrm{O}$ at product side by adding $2$ before ${\mathrm{KNO}}_3$.
$\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2+\mathrm{KI}\stackrel{\mathrm{heat}}{\to }{\mathrm{PbI}}_2+2{\mathrm{KNO}}_3$

ii. Balance $\mathrm{K}$ and $\mathrm{I}$ at reactant side by adding $2$ before $\mathrm{KI}$.

$\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2+2\mathrm{KI}\stackrel{\mathrm{heat}}{\to }{\mathrm{PbI}}_2+2{\mathrm{KNO}}_3$

vi. Balanced equation is:
$\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\left(\mathrm{aq}\right)+2\mathrm{KI}\left(\mathrm{aq}\right)\stackrel{\mathrm{heat}}{\to }{\mathrm{PbI}}_2\left(\mathrm{s}\right)+2{\mathrm{KNO}}_3\left(\mathrm{aq}\right)$
Lead(II) nitrate Potassium iodide Lead iodide Potassium nitrate

(Yellow)

So it is a double displacement reaction.