
Nitrogen gas is prepared:
(A) By heating ammonium nitrate
(B) By reacting chlorine with liquor ammonia
(C) By passing $\text{HN}{{\text{O}}_{\text{3}}}$ vapours on red hot copper
(D) By heating lead nitrate.
Answer
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Hint: Nitrogen gas is a colourless odourless gas. It is non-flammable and its nature is almost like inert gas but not truly inert. Here among the given options, nitrogen is prepared in the process where a metal is oxidised into its corresponding metal oxide and here the colour of the oxide is black. This process evolves nitrogen as another product.
Complete step by step solution:
We have been given four reactions in the options and each of these reactions produce a gas. We have to identify which reaction among these produces nitrogen.
Let’s look into each of the options carefully and find out their products.
-Option (A) When ammonium nitrate$\left( \text{N}{{\text{H}}_{\text{4}}}\text{N}{{\text{O}}_{3}} \right)$ is heated Nitrous oxide $\left( \text{ }{{\text{N}}_{\text{2}}}\text{O} \right)$ gas and water vapours are produced.
\[\underset{\text{Ammonium Nitrate}}{\mathop{\text{N}{{\text{H}}_{\text{4}}}\text{N}{{\text{O}}_{3}}}}\,\xrightarrow{\text{ }\!\!\Delta\!\!\text{ }}\text{ }\underset{\text{Nitrous Oxide}}{\mathop{{{\text{N}}_{\text{2}}}\text{O}\uparrow }}\,\text{+}\underset{\text{Water}}{\mathop{\text{ 2}{{\text{H}}_{2}}O}}\,\]
Thus we can see from this reaction that nitrogen is not produced.
Therefore, Option (A) is not the correct one.
-Option (B) The reaction of chlorine and ammonia can be of two types. One is when excess ammonia reacts with chlorine and another is when excess chlorine reacts with ammonia. Let’s look into each of the reactions.
(i) When excess ammonia reacts with chlorine, nitrogen gas along with ammonium chloride is produced.
\[\underset{\text{Ammonia}}{\mathop{8N{{H}_{3}}}}\,+\underset{\text{Chlorine}}{\mathop{3C{{l}_{2}}}}\,\to \underset{\text{Nitrogen}}{\mathop{{{N}_{2}}\uparrow }}\,+\underset{\text{Ammonium Chloride}}{\mathop{6N{{H}_{4}}Cl}}\,\]
(ii) On the other hand, when excess chlorine reacts with ammonia, nitrogen chloride and hydrogen chloride are produced.
\[\underset{\text{Ammonia}}{\mathop{N{{H}_{3}}}}\,+\underset{\text{Excess Chlorine}}{\mathop{3C{{l}_{2}}}}\,\to \underset{\text{Nitrogen Chloride}}{\mathop{NC{{l}_{3}}}}\,+\underset{\text{Hydrogen Chloride}}{\mathop{3HCl}}\,\]
As we can see that there are two possibilities for the reaction of ammonia with chlorine and it is not mentioned in the question whether ammonia or chlorine is in excess. So, we cannot choose this option as the correct one.
Therefore, option (B) is not the correct one.
-Option (C) When the concentrated $HN{{O}_{3}}$ vapour is passed over red hot copper, copper is oxidised by concentrated nitric acid thereby producing copper oxide whose colour is black. Nitrogen gas and water vapour are also produced in this process.
\[\underset{\text{Copper (Red Hot)}}{\mathop{5Cu}}\,+\underset{\text{Nitric acid (Conc}\text{.)}}{\mathop{2HN{{O}_{3}}}}\,\to \underset{\text{Copper Oxide}}{\mathop{5CuO}}\,+\underset{\text{Water}}{\mathop{{{H}_{2}}O}}\,+\underset{\text{Nitrogen}}{\mathop{{{N}_{2}}\uparrow }}\,\]
Hence we can clearly say that this reaction produces nitrogen gas and therefore option (C) is the correct one.
-When we heat lead(II) nitrate, yellow coloured lead (II) oxide is produced along with oxygen and nitrogen dioxide. The chemical equation is,
\[\underset{\text{Lead Nitrate}}{\mathop{2Pb{{\left( N{{O}_{3}} \right)}_{2}}}}\,\xrightarrow{\Delta }\underset{\text{Lead Oxide}}{\mathop{2PbO}}\,+\underset{\text{Nitrogen dioxide}}{\mathop{4N{{O}_{2}}\uparrow }}\,+\underset{Oxygen}{\mathop{{{O}_{2}}\uparrow }}\,\]
Therefore, option (D) is not the correct option.
Hence, from the above discussions, we can say that option (C) is the correct one.
Note: You should keep in mind that the reaction between copper and nitric acid produces different products in different conditions. When cold dilute nitric acid is passed over copper, it produces nitric oxide but when hot and concentrated nitric acid is passed over copper, nitrogen dioxide is produced. The reactions are given below:
\[\underset{\text{Copper}}{\mathop{3Cu}}\,+\underset{\text{Nitric acid (Cold dil}\text{.)}}{\mathop{8HN{{O}_{3}}}}\,\to \underset{\text{Copper Nitrate}}{\mathop{3Cu{{\left( N{{O}_{3}} \right)}_{2}}}}\,+\underset{\text{Water}}{\mathop{4{{H}_{2}}O}}\,+\underset{\text{Nitric oxide}}{\mathop{2NO\uparrow }}\,\]
\[\underset{\text{Copper}}{\mathop{Cu}}\,+\underset{\text{Nitric acid (Hot conc}\text{.)}}{\mathop{4HN{{O}_{3}}}}\,\to \underset{\text{Copper Nitrate}}{\mathop{Cu{{\left( N{{O}_{3}} \right)}_{2}}}}\,+\underset{\text{Water}}{\mathop{2{{H}_{2}}O}}\,+\underset{\text{Nitrogen dioxide}}{\mathop{2N{{O}_{2}}\uparrow }}\,\]
Remember, only when a red hot tube of copper is passed over hot and concentrated nitric acid, nitrogen gas is produced.
Complete step by step solution:
We have been given four reactions in the options and each of these reactions produce a gas. We have to identify which reaction among these produces nitrogen.
Let’s look into each of the options carefully and find out their products.
-Option (A) When ammonium nitrate$\left( \text{N}{{\text{H}}_{\text{4}}}\text{N}{{\text{O}}_{3}} \right)$ is heated Nitrous oxide $\left( \text{ }{{\text{N}}_{\text{2}}}\text{O} \right)$ gas and water vapours are produced.
\[\underset{\text{Ammonium Nitrate}}{\mathop{\text{N}{{\text{H}}_{\text{4}}}\text{N}{{\text{O}}_{3}}}}\,\xrightarrow{\text{ }\!\!\Delta\!\!\text{ }}\text{ }\underset{\text{Nitrous Oxide}}{\mathop{{{\text{N}}_{\text{2}}}\text{O}\uparrow }}\,\text{+}\underset{\text{Water}}{\mathop{\text{ 2}{{\text{H}}_{2}}O}}\,\]
Thus we can see from this reaction that nitrogen is not produced.
Therefore, Option (A) is not the correct one.
-Option (B) The reaction of chlorine and ammonia can be of two types. One is when excess ammonia reacts with chlorine and another is when excess chlorine reacts with ammonia. Let’s look into each of the reactions.
(i) When excess ammonia reacts with chlorine, nitrogen gas along with ammonium chloride is produced.
\[\underset{\text{Ammonia}}{\mathop{8N{{H}_{3}}}}\,+\underset{\text{Chlorine}}{\mathop{3C{{l}_{2}}}}\,\to \underset{\text{Nitrogen}}{\mathop{{{N}_{2}}\uparrow }}\,+\underset{\text{Ammonium Chloride}}{\mathop{6N{{H}_{4}}Cl}}\,\]
(ii) On the other hand, when excess chlorine reacts with ammonia, nitrogen chloride and hydrogen chloride are produced.
\[\underset{\text{Ammonia}}{\mathop{N{{H}_{3}}}}\,+\underset{\text{Excess Chlorine}}{\mathop{3C{{l}_{2}}}}\,\to \underset{\text{Nitrogen Chloride}}{\mathop{NC{{l}_{3}}}}\,+\underset{\text{Hydrogen Chloride}}{\mathop{3HCl}}\,\]
As we can see that there are two possibilities for the reaction of ammonia with chlorine and it is not mentioned in the question whether ammonia or chlorine is in excess. So, we cannot choose this option as the correct one.
Therefore, option (B) is not the correct one.
-Option (C) When the concentrated $HN{{O}_{3}}$ vapour is passed over red hot copper, copper is oxidised by concentrated nitric acid thereby producing copper oxide whose colour is black. Nitrogen gas and water vapour are also produced in this process.
\[\underset{\text{Copper (Red Hot)}}{\mathop{5Cu}}\,+\underset{\text{Nitric acid (Conc}\text{.)}}{\mathop{2HN{{O}_{3}}}}\,\to \underset{\text{Copper Oxide}}{\mathop{5CuO}}\,+\underset{\text{Water}}{\mathop{{{H}_{2}}O}}\,+\underset{\text{Nitrogen}}{\mathop{{{N}_{2}}\uparrow }}\,\]
Hence we can clearly say that this reaction produces nitrogen gas and therefore option (C) is the correct one.
-When we heat lead(II) nitrate, yellow coloured lead (II) oxide is produced along with oxygen and nitrogen dioxide. The chemical equation is,
\[\underset{\text{Lead Nitrate}}{\mathop{2Pb{{\left( N{{O}_{3}} \right)}_{2}}}}\,\xrightarrow{\Delta }\underset{\text{Lead Oxide}}{\mathop{2PbO}}\,+\underset{\text{Nitrogen dioxide}}{\mathop{4N{{O}_{2}}\uparrow }}\,+\underset{Oxygen}{\mathop{{{O}_{2}}\uparrow }}\,\]
Therefore, option (D) is not the correct option.
Hence, from the above discussions, we can say that option (C) is the correct one.
Note: You should keep in mind that the reaction between copper and nitric acid produces different products in different conditions. When cold dilute nitric acid is passed over copper, it produces nitric oxide but when hot and concentrated nitric acid is passed over copper, nitrogen dioxide is produced. The reactions are given below:
\[\underset{\text{Copper}}{\mathop{3Cu}}\,+\underset{\text{Nitric acid (Cold dil}\text{.)}}{\mathop{8HN{{O}_{3}}}}\,\to \underset{\text{Copper Nitrate}}{\mathop{3Cu{{\left( N{{O}_{3}} \right)}_{2}}}}\,+\underset{\text{Water}}{\mathop{4{{H}_{2}}O}}\,+\underset{\text{Nitric oxide}}{\mathop{2NO\uparrow }}\,\]
\[\underset{\text{Copper}}{\mathop{Cu}}\,+\underset{\text{Nitric acid (Hot conc}\text{.)}}{\mathop{4HN{{O}_{3}}}}\,\to \underset{\text{Copper Nitrate}}{\mathop{Cu{{\left( N{{O}_{3}} \right)}_{2}}}}\,+\underset{\text{Water}}{\mathop{2{{H}_{2}}O}}\,+\underset{\text{Nitrogen dioxide}}{\mathop{2N{{O}_{2}}\uparrow }}\,\]
Remember, only when a red hot tube of copper is passed over hot and concentrated nitric acid, nitrogen gas is produced.
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