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Introduction to Le Chatelier's Principle

0.5 moles of ...

Question

$0.5$ moles of $B a C l_{2}$ is mixed with $0.2$ moles of $(N H_{4})_{3} P O_{4}$ . Reactant present in excess (with extra number of moles of that reactant) is :

A

B a C l_{2}, 0.3 m o l

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B

(N H_{4})_{3} P O_{4}, 0.2 m o l

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C

(N H_{4})_{3} P O_{4}, 0.1 m o l

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D

B a C l_{2}, 0.2 m o l

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Solution

The correct option is D $B a C l_{2}, 0.2 m o l$
$3 B a C l_{2} + 2 (N H_{4})_{3} P O_{4} ⟶ B a_{3} (P O_{4})_{2} + 6 N H_{4} C l$
$3 m o l 2 m o l 1 m o l$
Given: $0.5 m o l 0.2 m o l 0.1 m o l$
Thus, $(N H_{4})_{3} P O_{4}$ is the limiting reactant giving $0.1$ mol $B a_{3} (P O_{4})_{2}$ .
$B a C l_{2}$ required by $0.2$ mol of $(N H_{4})_{3} P O_{4} = 0.3 m o l$ .
Thus, $B a C l_{2}$ in excess $= 0.5 - 0.3 = 0.2 m o l$

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