Equivalent weight of NH 3 in the change N 2- NH 3 isA- 17B- 17-2C- 17-3D- 17-6

The correct option is C 17/30N_2→ 3NH_3 “ to balance charge N_2 +6e^ → NH_3 (OR) 1/2N_2 +3e^ → NH_3 Melecular mass of NH_3 =17 Equivalent weight of NH_3 =17/ ...

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Byju's Answer

Standard XII

Chemistry

Enthalpy of Neutralisation

Equivalent we...

Question

Equivalent weight of $N H_{3}$ in the change $N_{2} \Rightarrow N H_{3}$ is

\frac{17}{6}

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17

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\frac{17}{2}

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\frac{17}{3}

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Solution

The correct option is D $\frac{17}{3}$
$_{2} \to - 3 N H_{3}$ “ to balance charge
$N_{2} + 6 e^{-} \to N H_{3}$
(OR)
$\frac{1}{2} N_{2} + 3 e^{-} \to N H_{3}$
Melecular mass of $N H_{3} = 17$
Equivalent weight of $N H_{3} = \frac{17}{3}$

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Similar questions

Q. Assertion :Equivalent weight of

N H_{3}

in the reaction ,

N_{2} \to N H_{3}

is 17/3 while that of

N_{2}

is 28/6. Reason: Equivalent weight

= \frac{Molecular weight}{Number of electrons lost or gained per molecule}

N_{2} + 3 H_{2} ⟶ 2 N H_{3}

The equivalent weight of

N H_{3} = \frac{17}{x}

. The value of

x

is:

Q. Calculate the Equivalent weight of

N H_{3}

in the change

N_{2} \to N H_{3}

Question 9.
Sum of $\frac{4}{17}$ and $\frac{15}{17}$ is
(a) $\frac{19}{17}$
(b) $\frac{11}{17}$
(c) $\frac{19}{34}$
(d) $\frac{2}{17}$

Q. Equivalent weight of

{N H}_{3}

in the change

N_{2} \to {N H}_{3}

is:

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Equivalent weight of NH3 in the change N2⇒NH3 is

The correct option is D 1730N2→−3NH3 “ to balance charge N2+6e−→NH3 (OR) 12N2+3e−→NH3 Melecular mass of NH3=17 Equivalent weight of NH3=173

Equivalent weight of $N H_{3}$ in the change $N_{2} \Rightarrow N H_{3}$ is

The correct option is D $\frac{17}{3}$
$_{2} \to - 3 N H_{3}$ “ to balance charge
$N_{2} + 6 e^{-} \to N H_{3}$
(OR)
$\frac{1}{2} N_{2} + 3 e^{-} \to N H_{3}$
Melecular mass of $N H_{3} = 17$
Equivalent weight of $N H_{3} = \frac{17}{3}$