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Oxyacids of Phosphorus

Equivalent ma...

Question

Equivalent masses of $N H_{3}$ in the reactions are :
(i) $4 N H_{3} + 5 O_{2} ⟶ 4 N O + 6 H_{2} O$
(ii) $2 N H_{3} ⟶ N_{2} + 3 H_{2}$

5 : 6

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6 : 5

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5 : 3

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3 : 5

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Solution

The correct option is C $3 : 5$
$4 N H_{3} + 5 O_{2} ⟶ 4 N O + 6 H_{2} O$ ,
-3 +2
So, eq. wt. of $N H_{3} = \frac{17}{5}$ .
$2 N H_{3} ⟶ N_{2} + 3 H_{2}$
-3 0
So, eq. wt. of $N H_{3} = \frac{17}{3}$ .
So, their ratio is $3 : 5$ .

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