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 milli mol of $P b C l_{2}$ is mixed with 50 milli mol of ${(N H_{4})}_{3} P O_{4},$ then the amount of $P b_{3} {(P O_{4})}_{2}$ formed is ___________milli mol (nearest integer)

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$3 P b C l_{2} + 2 {(N H_{4})}_{3} P O_{4} \to P b_{3} {(P O_{4})}_{2} + 6 N H_{4} C l$

$3 m m o l e 2 m m o l e 1 m m o l e$

$72 m m o l e 50 m m o l e ?$

$(\begin{array}{l} 48 m m o l e \\ r e q u i r e d \end{array})$

Limiting Reagent is ${(N H_{4})}_{3} P O_{4}$

$∴$ 2m mol of

${(N H_{4})}_{3} P O_{4} \to 1 m m o l o f P b_{3} {(P O_{4})}_{2}$

$48 m m o l o f {(N H_{4})}_{3} P O_{4} \to ?$

No.of m mole of $P b_{3} {(P O_{4})}_{2} = \frac{48 \times 1}{2}$

$= 24 m m o l e o f P b_{3} {(P O_{4})}_{2}$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE