Question

The questions given below are related to the manufacturing of ammonia.

(i) Name the process.

(ii) In what ratio must the reactants be taken.

(iii) Name the catalyst used.

(iv) Give the equation for the manufacturing of ammonia.

(v) Ammonia can act as reducing agent, write a relevant equation for such a reaction.

Answer 1

Manufacturing of Ammonia$\left({\mathrm{NH}}_3\right)$

(i) Name of Process

• Ammonia$\left({\mathrm{NH}}_3\right)$ is manufactured by Haber's process.

(ii) Ratio of reactants in manufacturing of Ammonia

• Nitrogen$\left({\mathrm{N}}_2\right)$ and Hydrogen$\left({\mathrm{H}}_2\right)$ are the two reactants in the formation of Ammonia.
• These are taken in ratio 1:3 by volume respectively.

(iii) Name of catalyst used

• Finely divided Iron$\left(\mathrm{Fe}\right)$ is used as a catalyst.
• Small amount of promoter Molybdenum$\left(\mathrm{Mo}\right)$ is also used with the catalyst to enhance the efficiency of the catalyst.

(iv) Equation for manufacturing of Ammonia

• ${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\stackrel{450-{500}^{\mathrm{o}}\mathrm{C}}{\rightleftharpoons }2{\mathrm{NH}}_3\left(\mathrm{g}\right)+\mathrm{Heat}$

(Nitrogen) (Hydrogen) (Ammonia)

(v) Ammonia as a reducing agent

• Chlorine$\left({\mathrm{Cl}}_2\right)$ gas is reduced to Ammonium chloride$\left({\mathrm{NH}}_4\mathrm{Cl}\right)$ in presence of excess Ammonia$\left({\mathrm{NH}}_3\right)$.
• First Chlorine reacts with Ammonia to give Hydrogen chloride$\left(\mathrm{HCl}\right)$ which further reacts with excess Ammonia and produces Ammonium chloride.
• The overall reaction is as follows:
• $8{\mathrm{NH}}_3\left(\mathrm{g}\right)+3{\mathrm{Cl}}_2\left(\mathrm{g}\right)\to {\mathrm{N}}_2\left(\mathrm{g}\right)+6{\mathrm{NH}}_4\mathrm{Cl}\left(\mathrm{s}\right)$