Question

The n factor for $N{H_3}$ ​ in the reaction:
${N_2} + 3{H_2} \to 2N{H_3}$
A. 2
B. 3
C. 1
D. 4

Answer 1

Initially you must be aware of the reaction which takes places with nitrogen and hydrogen. Now, after getting the knowledge of the kind of reaction then it could be easy to calculate the n factor of the reaction.

Complete step by step answer:
Ammonia may be a nutritious, inorganic nitrogen compound that's found naturally within the air, soil, water and living tissues. A stable binary hydride, and the simplest pnictogen hydride, ammonia is a colourless gas.
At room temperature, it is a gas. Its formula is $N{H_3}$ within the gaseous phase and when dissolved in water, much of it ionizes to $N{H_4}^ +$ called ammonium ion.
Losing electrons is called oxidation while gaining of electrons is called reduction.
This is an oxidation-reduction (redox) reaction:

{6\;{H^0}\; - \;6\;{e^ - }\; \to \;6\;{H^{ + 1}}{\text{ }}\;} \\\ {2\;{N^0}\; + \;6\;{e^ - }\; \to \;2\;{N^{ - 3}}\;{\text{ }}} \end{array}$$ $${H_2}$$ is a reducing agent, $${N_2}$$ is an oxidizing agent. n-factor is the number of moles of electrons given or gained. Since this is an acid base reaction in which $$2{H^ + }$$ is removed from hydro sulphuric acid and the number of replaceable hydrogens is $$2{H^ + }$$ . Thus for 2 moles of $$N{H_3}$$ 2 moles of $${H^ + }$$ is added. For 1 mole of $$N{H_3}$$ . $${H^ + }$$mole is added. NH3 accepts $${H^ + }$$ion. So, its N-factor is 1. **So, the correct answer is Option C.** **Note:** A redox reaction may be a sort of reaction that involves a transfer of electrons between two species in which both oxidation and reduction occurs. An oxidation-reduction reaction is any chemical reaction in which the oxidation number of a molecule, atom, or ion change by gaining or losing an electron.