Consider the following reaction: 3 PbCl _ 2 + 2 ( NH _ 4 ) _ 3 PO _...

The reaction is given as,

$3{PbCl}_{2} + 2{\left({NH}_{4}\right)}_{3}{PO}_{4} \to {Pb}_{3}{\left({PO}_{4}\right)}_{2} + 6{NH}_{4}Cl$

The products formed in the above reaction is ${Pb}_{3}{\left({PO}_{4}\right)}_{2}\text{and}{NH}_{4}Cl$ .

According to the above stoichiometric equation, three mole of each lead chloride and two moles of ${\left({NH}_{4}\right)}_{3}{PO}_{4}$ is required to make one mole of ${Pb}_{3}{\left({PO}_{4}\right)}_{2}\text{and}6\text{moles of}{NH}_{4}Cl$

Here the Limiting Reagent is ${PbCl}_{2}$

$mmol$ of ${Pb}_{3} ( {PO}_{4}{)}_{2}$ formed

$= \dfrac{mmol\text{of}{PbCl}_{2}\text{reacted}}{3} = \dfrac{72}{3}$

$= 24m mol$