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Equivalent Mass

Mg can reduce...

Question

$M g$ can reduce $N O_{3}^{-}$ to $N H_{3}$ in basic medium.
$N O_{3}^{-} (a q) + M g (s) + H_{2} O \to M g (O H)_{2} (s) + O H^{-} (a q) + N H_{3} (g)$
A $25.0 m L$ sample of $N O_{3}^{-}$ solution was treated with $M g (s)$ . The $N H_{3} (g)$ was passed into $100 m L$ of $0.15 N H C l$ .
The excess of $H C l$ required $32.10 m L$ of $0.10 N N a O H$ for neutralization. What was the molarity of $N O_{3}^{-}$ ions in the original sample?

A

0.47125

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B

0.3025

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C

3.025

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D

0.5005

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Solution

The correct option is A $0.47125$
$Milliequivalents of N H_{3} formed = Milliequivalents of H C l used = 100 \times 0.15 - 32.10 \times 0.10 = 11.79$

Here, n-factor of $N H_{3}$ is $1$ (acid - base reaction).

For redox change,

$∴ Milliequivalents of N H_{3} for n-factor 8 = 8 \times 11.79 = 94.32$
$∴ Normality of N O_{3}^{-} = \frac{94.32}{25} = 3.77$
$Molarity of N O_{3}^{-} = \frac{3.77}{8} = 0.47125$

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Similar questions

M g

N O_{3}^{-}

N H_{3}

in basic medium.

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

25.0 m L

N O_{3}^{-}

solution was treated with

M g (s)

N H_{3} (g)

was passed into

100 m L

0.15 N H C l

.
The excess of

H C l

32.10 m L

0.10 N N a O H

for neutralization. What was the molarity of

N O_{3}^{-}

ions in the original sample?

Q. 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

N H_{3} (g)

was passed into

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0.1 M

H C l

H C l

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Q. ways in which the equilibrium concentration of

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N a^{+}

C l^{-}

A g^{+}

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A g C l .

N a^{+} (a q) + C l^{-} (a q) + A g^{+} (a q) + N O_{3}^{-} (a q) ⇌ A g C l (s) + N a^{+} (a q) + N O_{3}^{-} (a q) Δ H = - 65.9 k J

M g + (N O_{3})^{-} \to

M g + (S O_{4})^{2 -} \to

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