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Mole Concept

In the reacti...

Question

In the reaction given,
$M g (s) + N O_{3}^{⊝} + H_{2} O \to M g (O H)_{2} (s) + ⊝ O H (a q) + N H_{3} (g)$
$20$ mL of sample of $N O_{3}^{⊝}$ solution is treated with $M g$ . The $N H_{3} (g)$ was passed into $50$ mL of $0.1 M$ $H C l$ . The excess $H C l$ required $30$ mL of $0.1 M$ $K O H$ for its neutralization, then the molarity of $N O_{3}^{⊝}$ ions in the original sample will be :

0.01 M

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1 M

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1.01 M

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0.1 M

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Solution

The correct option is D $0.1 M$
Balance the equation in basic medium.
$/ 8 e^{-} + 6 H_{2} O + N O_{3}^{⊝} \to N H_{3} + 9 ⊝ O H$
$2 ⊝ O H + M g \to 2 + M g (O H)_{2} + / 2 e^{-}] \times 4$
____________________________________________
$6 H_{2} O + N O_{3}^{⊝} + 4 M g \to N H_{3} + 4 M g (O H)_{2} + ⊝ O H$
____________________________________________
Note: The milliequivalents of $N H_{3}$ is derived using valence factor of $N H_{3} = 1$ (in acid base reaction).
In redox reaction, valence factor for $N H_{3}$ is $8$ .
Milliequivalents of $N H_{3}$ formed $=$ Milliequivalents of $H C l$ used for $N H_{3}$
$= (50 \times 0.1 \times 1) - (30 \times 0.1 \times 1)$ $= 2$ meq.
Thus, meq.of $N H_{3}$ for valence factor of $8 = 8 \times 2 = 16$
Also, meq. of $N O_{3}^{⊝} =$ mEq of $N H_{3}$ $= 8 \times 2 = 16$ meq.
$N_{N O_{3}^{⊝}} \times V_{m L} = 16$
$∴ N_{N O_{3}^{⊝}} \times 20 = 16$
$N_{N O_{3}^{⊝}} = \frac{16}{20} = 0.8 M$
$M_{N O_{3}^{⊝}} = \frac{0.8}{8} = 0.1 M$

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Similar questions

M g

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N O_{3}^{-}

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in basic solution as follows:

N O_{3}^{-} + M g (s) + H_{2} O ⟶ M g (O H)_{2} (s) + O H^{-} (a q) + N H_{3} (g)

25.0

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Q. One gram of commercial

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