$0.5$ moles of $B a C l_{2}$ is mixed with $0.2$ moles of $(N H_{4})_{3} P O_{4}$ . Maximum number of moles of barium phosphate formed in this reaction is :

0.1

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0.2

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0.3

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0.5

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Solution

The correct option is A $0.1$
The balanced reaction is as follows:
$3 B a C l_{2} + 2 (N H_{4})_{3} P O_{4} ⟶ B a_{3} (P O_{4})_{2} + 6 N H_{4} C l$
$3 m o l$ $2 m o l$ $1 m o l$
Given: $0.5 m o l$ $0.2 m o l$
Thus, $(N H_{4})_{3} P O_{4}$ is the limiting reactant giving $0.1$ mol $B a_{3} (P O_{4})_{2}$ .

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