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Introduction

Ag N H 3 2 +...

Question

$A g {(N H_{3})}_{2}^{+} \to {A g}^{+} + 2 N H_{3}$
How many moles of $A g$ are required for the production of $11$ moles of ammonia $(N H_{3})$ ?

7.5 m o l

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8.5 m o l

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5.5 m o l

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9.5 m o l

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Solution

The correct option is C $5.5 m o l$
$A g (N H_{3})_{2}^{+} \to A g^{+} + 2 N H_{3}$
It is clear that, $2$ moles of $N H_{3}$ are produced by $1$ mol of $A g (N H_{3})_{2}^{+}$ , so 2 moles of $N H_{3}$ required $1$ mol of $A g$ .
Now, as per question, to determine the number of moles of $A g$ required for the proudction of $11$ moles of ammonia.
For $2$ moles of $N H_{3}$ , $1$ mol of $A g$ is required.
so for $11$ moles of $N H_{3}$ , moles of $A g$ required = $11 \times (1 / 2) = 5.5 m o l e s$

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Similar questions

A g (N H_{3})_{2}^{+}

. . . A g (N H_{3})_{2}^{+} \to . . . A g^{+} + . . . N H_{3}

A g {(N H_{3})}_{2}^{+} \to {A g}^{+} + 2 N H_{3}

5.8 g

A g {(N H_{3})}_{2}^{+}

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{N H}_{3}

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0.1 M

A g {N O}_{3}

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2 \times 10^{- 7} M

{[A g {({N H}_{3})}_{2}]}^{+} ⇌ {A g}^{+} + 2 {N H}_{3}

K_{d i s s .} = \frac{[{A g}^{+}] {[{N H}_{3}]}^{2}}{[A g {({N H}_{3})}_{2}^{+}]} = 6.8 \times 10^{- 8}

N_{2}

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0.2 M

[A g {({N H}_{3})}_{2}] {N O}_{3}

A g {({N H}_{3})}_{2}^{+} ⇌ {A g}^{+} + 2 {N H}_{3}

K_{d i s s .} = \frac{[{A g}^{+}] {[{N H}_{3}]}^{2}}{[A g {({N H}_{3})}_{2}^{+}]} = 6.8 \times 10^{- 8}

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