Consider the following reaction, If 72 mmol of , is mixed with 50 mmol of , then the amount of formed is _______ mmol (nearest integer).

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Solution

Q.
Consider the following reaction:

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

If 72 mmol of $P b C l_{2}$ is mixed with 50 mmol of ${(N H_{4})}_{3} P O_{4}$ , then the amount of $P b_{3} {(P O_{4})}_{2}$ formed is _______ mmol (nearest integer). [2024]

Ans.
(24)    $3 {PbCl}_{2} + 2 {({NH}_{4})}_{3} {PO}_{4} \to {Pb}_{3} {({PO}_{4})}_{2} + 6 {NH}_{4} Cl$

  Finding limiting reagent:

   $\frac{n_{{PbCl}_{2}}}{Stoichiometric coefficient of {PbCl}_{2}} = \frac{72}{3} = 24$

$\frac{n_{{({NH}_{4})}_{3} P O_{4}}}{Stoichiometric coefficient of {({NH}_{4})}_{3} P O_{4}} = \frac{50}{2} = 25$

  As 24 < 25, $P b C l_{2}$ is limiting reagent. Amount of product i.e. $P b_{3} {(P O_{4})}_{2}$ formed depends upon amount of limiting reagent i.e. $P b C l_{2}$ :

By stoichiometry:

$\frac{n_{{Pb}_{3} {({PO}_{4})}_{2}}}{1} = \frac{n_{{PbCl}_{2}}}{3}$

   $\frac{n_{{Pb}_{3} {({PO}_{4})}_{2}}}{1} = \frac{72 mmol}{3}$

   $n_{{Pb}_{3} {({PO}_{4})}_{2}} = 24 mmol$

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