Question

In the reaction,
$4HN{{O}_{3}}+{{P}_{4}}{{O}_{10}}\to 4HP{{O}_{3}}+X$ ; the product 'X' is:
(a)- ${{N}_{2}}{{O}_{5}}$
(b)- ${{N}_{2}}{{O}_{3}}$
(c)- $N{{O}_{2}}$
(d)- ${{H}_{2}}O$

Answer 1

In the reaction given above $HN{{O}_{3}}$ is known as nitric acid, and ${{P}_{4}}{{O}_{10}}$ is phosphorus pentoxide there will be the formation of metaphosphoric acid and a compound of nitrogen in which the oxidation state of nitrogen will be +5.

Complete Solution :
- Nitric acid having the formula $HN{{O}_{3}}$ is a compound of a nitrogen atom, and it is a very strong acid. ${{P}_{4}}{{O}_{10}}$ is a compound of phosphorus and its name is phosphorus pentoxide. It is one of the main oxides of phosphorus.
- When the phosphorus pentoxide reacts with water it forms phosphoric acid having formula ${{H}_{3}}P{{O}_{4}}$, this is because of the fact that phosphorus pentoxide is acidic in nature. So, because of this affinity towards water, ${{P}_{4}}{{O}_{10}}$ is a powerful dehydrating agent. Therefore, it can extract many organic and inorganic compounds.
- In the reaction given above $HN{{O}_{3}}$ is known as nitric acid, and ${{P}_{4}}{{O}_{10}}$ is phosphorus pentoxide there will be the formation of metaphosphoric acid and a compound of nitrogen in which the oxidation state of nitrogen will be +5.

So the complete reaction will be:
$4HN{{O}_{3}}+{{P}_{4}}{{O}_{10}}\to 4HP{{O}_{3}}+2{{N}_{2}}{{O}_{5}}$
Hence, 4 moles of nitric acid and 1 mole of phosphorus pentoxide react to form 4 moles of metaphosphoric acid and 2 moles of dinitrogen pentoxide. And in Dinitrogen pentoxide, the oxidation state of nitrogen is +5. So, the correct answer is “Option A”.

Note: With other acid the reaction of phosphorus pentoxide are given below:
- With sulfuric acid the reaction will be:
$2{{H}_{2}}S{{O}_{4}}+{{P}_{4}}{{O}_{10}}\to 2S{{O}_{3}}+4HP{{O}_{3}}$
- With perchloric acid the reaction will be:
$4HCl{{O}_{4}}+{{P}_{4}}{{O}_{10}}\to 2C{{l}_{2}}{{O}_{7}}+4HP{{O}_{3}}$
- With acetamide the reaction will be:
$2C{{H}_{3}}CON{{H}_{2}}+{{P}_{4}}{{O}_{10}}\to 2C{{H}_{3}}CN+4HP{{O}_{3}}$