Consider the following reaction: $3PbC{l}_2+2{\left(N{H}_4\right)}_{3}P{O}_4\to P{b}_3{\left(P{O}_4\right)}_2+6N{H}_{4}Cl$

If 72 milli mol of $PbC{l}_2$ is mixed with 50 milli mol of ${\left(N{H}_4\right)}_{3}P{O}_4,$ then the amount of $P{b}_3{\left(P{O}_4\right)}_2$  formed is ___________milli mol (nearest integer)

$3PbC{l}_2+2{\left(N{H}_4\right)}_{3}P{O}_4\to P{b}_3{\left(P{O}_4\right)}_2+6N{H}_{4}Cl$

$3mmole 2mmole 1mmole$

$72mmole 50mmole?$

                               $\left(\begin{array}{l}48mmole\\ required\end{array}\right)$

Limiting Reagent is  ${\left(N{H}_4\right)}_{3}P{O}_4$

$\therefore$  2m               mol               of 

${\left(N{H}_4\right)}_{3}P{O}_4\to 1mmolofP{b}_3{\left(P{O}_4\right)}_2$

$48mmolof{\left(N{H}_4\right)}_{3}P{O}_4\to ?$

No.of m mole of  $P{b}_3{\left(P{O}_4\right)}_2=\frac{48\times 1}{2}$

$=24mmoleofP{b}_3{\left(P{O}_4\right)}_2$