Lead($\mathrm{II}$) nitrate reacts with potassium iodide to form lead($\mathrm{II}$) iodide and potassium nitrate. How do you write a balance equation for this?

Steps for balancing an equation:

• Creating a chemical equation that is not balanced
• On the reactant and product sides, the atoms must be balanced.
• In a chemical equation, coefficients are used to balance the mass.

Formation of Lead iodide and Potassium nitrate:

• Writing an unbalanced equation where Lead$\left(\mathrm{II}\right)$ nitrate reacting with Potassium iodide to give lead$\left(\mathrm{II}\right)$ iodide and Potassium nitrate. $\underset{\mathrm{Lead}\mathrm{nitrate}}{\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\left(\mathrm{l}\right)}+\underset{\mathrm{Potassium}\mathrm{iodide}}{\mathrm{KI}\left(\mathrm{s}\right)}\to \underset{\mathrm{Lead}\mathrm{iodide}}{{\mathrm{PbI}}_2\left(\mathrm{s}\right)}+\underset{\mathrm{Potassium}\mathrm{nitrate}}{{\mathrm{KNO}}_3\left(\mathrm{l}\right)}$
• Writing the number of atoms on the reactant and product side

Atom	Reactant side	Product side
$\mathrm{Pb}$	$1$	$1$
$\mathrm{N}$	$2$	$1$
$\mathrm{O}$	$5$	$3$
$\mathrm{K}$	$1$	$1$
$\mathrm{I}$	$1$	$2$

• For balancing nitrogen we have to multiply KNO3 by 2. $\underset{\mathrm{Lead}\mathrm{nitrate}}{\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\left(\mathrm{l}\right)}+\underset{\mathrm{Potassium}\mathrm{iodide}}{\mathrm{KI}\left(\mathrm{s}\right)}\to \underset{\mathrm{Lead}\mathrm{iodide}}{{\mathrm{PbI}}_2\left(\mathrm{s}\right)}+\underset{\mathrm{Potassium}\mathrm{nitrate}}{2{\mathrm{KNO}}_3\left(\mathrm{l}\right)}$
• For balancing Iodine we need to multiply KI by 2. $\underset{\mathrm{Lead}\mathrm{nitrate}}{\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\left(\mathrm{l}\right)}+\underset{\mathrm{Potassium}\mathrm{iodide}}{2\mathrm{KI}\left(\mathrm{s}\right)}\to \underset{\mathrm{Lead}\mathrm{iodide}}{{\mathrm{PbI}}_2\left(\mathrm{s}\right)}+\underset{\mathrm{Potassium}\mathrm{nitrate}}{2{\mathrm{KNO}}_3\left(\mathrm{l}\right)}$
• After balancing Nitrogen and Iodine, Oxygen and lead are automatically balanced.

Therefore, the balanced equation for the reaction of lead nitrate and Potassium iodide is given as $\underset{\mathrm{Lead}\mathrm{nitrate}}{\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2}\left(\mathrm{l}\right)+\underset{\mathrm{Potassium}\mathrm{iodide}}{2\mathrm{KI}\left(\mathrm{s}\right)}\to \underset{\mathrm{Lead}\mathrm{iodide}}{{\mathrm{PbI}}_2\left(\mathrm{s}\right)}+\underset{\mathrm{Potassium}\mathrm{nitrate}}{2{\mathrm{KNO}}_3\left(\mathrm{l}\right)}$