Consider the following reaction:
$3 P b C l_{2} + 2 {(N H_{4})}_{3} P O_{4} \to P b_{3} {(P O_{4})}_{2} + 6 N H_{4} C l$
If 72 millimoles of $P b C l_{2}$ is mixed with 50 millimoles of ${(N H_{4})}_{3} P O_{4}$ , then the amount $P b_{3} {(P O_{4})}_{2}$ of formed is ___________ millimoles (nearest integer)

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answer is 24.

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Detailed Solution

$3 P b C l_{2} + 2 {(N H_{4})}_{3} P O_{4} \to P b_{3} {(P O_{4})}_{2} + 6 N H_{4} C l$
3m mole 2m mole 1m mole 72m mole 50m mole ?
$(\begin{array}{l} 48 m m o l e \\ r e q u i r e d \end{array})$
Limiting Reagent is ${(N H_{4})}_{3} P O_{4}$
$∴ 2 m m o l o f {(N H_{4})}_{3} P O_{4} \to 1 m m o l o f P b_{3} {(P O_{4})}_{2} ∴ 2 m m o l e o f {(N H_{4})}_{3} P O_{4} \to 1 m m o l o f P b_{3} {(P O_{4})}_{2} 48 m m o l o f {(N H_{4})}_{3} P O_{4} \to ?$
No.of m mole of $P b_{3} {(P O_{4})}_{2} = \frac{48 \times 1}{2}$
$= 24 m m o l e o f P b_{3} {(P O_{4})}_{2}$