Question

What compound of sulfur is obtained when conc. $HN{O_3}$ oxidizes ${S_8}$?

Answer 1

Octasulfur is an inorganic compound with the chemical formula ${S_8}$.it is a yellow solid, and is odorless and tasteless. It is the most common allotropes of sulfur. It is a major industrial chemical that occurs widely in nature.

Complete answer:
The concentrated nitric acid is a powerful oxidizing agent. It can oxidize Sulphur to sulfuric acid. So, the Sulphur is oxidizing from an oxidation state of $0$ to $+ 6$ and oxidation number of nitrogen in nitric acid is decreased from $+ 5$ to $+ 4$. Let’s consider ${S_8}$ loses six electrons per Sulphur atom, and forms sulfuric acid. So taking into account the acid nature of the medium, we balance ${S_8} \to 8{H_2}S{O_4}$ we get:
${S_8} + 32{H_2}O \to 8{H_2}S{O_4} + 48{H^ + } + 48{e^ - }$
Now, consider the reduction half-reaction of nitric acid decomposition to nitrogen dioxide. So, balancing $HN{O_3} \to N{O_2}$, we get
$HN{O_3} + {H^ + } \to N{O_2} + {H_2}O$
Combining these half-reactions to eliminate the electrons from the equations, we get
${S_8} + 48HN{O_3} \to 8{H_2}S{O_4} + 48N{O_2} + 16{H_2}O$
Sulfur is oxidized to sulfuric acid producing $8$ moles of sulfuric acid reacting with $48$ moles of nitric acid and nitric acid is reduced to $48$ moles of nitrogen dioxide and giving $16$ moles of water in the product as well.

Note:
Sulfur on reacting with concentrated nitric acid gives sulfuric acid instead of Sulphur dioxide because concentrated nitric acid acts as an oxidizing agent for both sulfur and carbon. In case of Sulphur, it will get the oxygen atoms and the water needed for the reaction:
$S + 6HN{O_3} \to {H_2}S{O_4} + 2{H_2}O + 6N{O_2}$
One mole of sulfur reacts with six nitric acid moles and produces one sulfuric acid, six nitrogen dioxide and two water molecules. So if sulfur is converted to $S{O_2}$ first, it will proceed into $S{O_3}$ and eventually into ${H_2}S{O_4}$.