Animated Solution for Chemistry - Redox Reactions: The amount of water produced (in g) in the oxidation of $1$ mole of rhombic sulphur by conc. ${\text{HNO}}_3$ to a compound with the highest oxidation state of sulphur is ______. (Given data : Molar mass of water = $18{\text{ g mol}}^{-1}$)

Let's decode the chemistry happening here. We are dealing with the oxidation of rhombic sulphur by concentrated nitric acid. Rhombic sulphur naturally exists as $S_8$ molecules. When it reacts with a strong oxidizing agent like concentrated $HN{O}_3$, it is forced into its highest possible oxidation state.

For sulphur, the highest oxidation state is $+6$, which corresponds to the formation of sulphuric acid ($H_{2}S{O}_4$).

Simultaneously, the concentrated nitric acid must undergo reduction. In such vigorous redox reactions, concentrated $HN{O}_3$ typically reduces to nitrogen dioxide ($N{O}_2$) gas. To balance the remaining hydrogen and oxygen atoms, water ($H_{2}O$) will be formed as a byproduct.

To balance this reaction, we need to look closely at the electron transfer. Let's start with the oxidation half-reaction. One molecule of $S_8$ contains eight sulphur atoms, all starting at an oxidation state of $0$.

In the product, $H_{2}S{O}_4$, each sulphur atom is at a $+6$ oxidation state. This means each sulphur atom loses $6$ electrons. For an entire $S_8$ molecule, the total loss is a massive $8\times 6=48$ electrons!

Now, let's examine the reduction half-reaction. The nitrogen in $HN{O}_3$ starts at an oxidation state of $+5$ and reduces to $+4$ in $N{O}_2$. This means each nitric acid molecule gains exactly $1$ electron.

For the redox process to be valid, the number of electrons lost must perfectly equal the number of electrons gained. Since the sulphur molecule loses $48$ electrons, we require exactly $48$ molecules of nitric acid to accept them.

This gives us the skeletal balanced equation for the main species:

Now for the final balancing act: hydrogen and oxygen. On the reactant side, we have $48$ hydrogen atoms from the nitric acid. On the product side, the $8$ molecules of sulphuric acid account for $16$ hydrogen atoms.

This leaves us with $48-16=32$ hydrogen atoms that need a home. These $32$ hydrogens will perfectly form $16$ molecules of water ($H_{2}O$).

Let's verify the oxygen atoms to be absolutely sure. On the left, we have $48\times 3=144$ oxygen atoms. On the right, we have $(8\times 4)+(48\times 2)+(16\times 1)=32+96+16=144$ oxygen atoms. It balances perfectly!

The complete balanced equation is:

The question specifically asks for the mass of water produced from exactly $1$ mole of rhombic sulphur.

Looking at our beautifully balanced equation, we can see the stoichiometric ratio directly: $1$ mole of $S_8$ yields exactly $16$ moles of $H_{2}O$.

Since the molar mass of water is given as $18{\text{ g mol}}^{-1}$, we simply multiply the number of moles by the molar mass: