Question

The amount of water produced (in g) in the oxidation of 1 mole rhombic Sulphur by conc. \[HN{O_3}\] to a compound with the highest oxidation state of Sulphur is ___.
(Given data: Molar mass of water = 18g \[mo{l^{ - 1}}\])

Answer 1

Hint: The rhombic Sulphur reacts with nitric acid to form sulphuric acid, nitrogen dioxide and water. To know the mole of water balances the whole equation. The number of moles present on the left side of reaction is the same as the number of moles present on the right side of reaction.

Complete step by step answer:
Molar mass of water is 18g/mol.
It is given that 1 mole of rhombic Sulphur is reacted with concentrated nitric acid. The formula of rhombic Sulphur is \[{S_8}\].
In this reaction, a redox reaction is taking place. The redox reaction is defined as a reaction where transfer of electrons takes place during the reaction to form the product where oxidation and reduction reaction takes place simultaneously.
In the oxidation reaction, loss of electrons takes place. In oxidation increase in oxidation state of the atom takes place from the reactant to the product.
In reduction reaction, gain of electrons takes place. In reduction, decrease in oxidation state of the atom takes place from the reactant to the product.
The reaction of 1 mole rhombic Sulphur with concentrated nitric acid is shown below.
\[S_8^0 + 48{H^{ + 5}}N{O_3} \to 8{H_2}{S^{ + 6}}{O_4} + 48{N^{ + 4}}{O_2} + 16{H_2}O\]
In this reaction, 1 mole of rhombic Sulphur reacts with 48 mole of nitric acid to give 8 mole of sulphuric acid, 48 mole of nitrogen dioxide and 16 mole of water.
In this reaction rhombic Sulphur is oxidized to sulphuric acid and nitric acid is reduced to nitrogen dioxide.
The formula to calculate the number of moles is shown below.
$n = \dfrac{m}{M}$
Where,
n is the number of moles.
m is the given mass
M is the molecular weight
The molecular weight of water is 18 g/mol
To calculate the mass, substitute the values of moles and molar mass in the above expression.
$ \Rightarrow 16mol = \dfrac{m}{{18g/mol}}$
$ \Rightarrow m = 16mol \times 18g/mol$
$ \Rightarrow m = 288g$

Therefore, the mass of water produced (in g) in the oxidation of 1 mole rhombic Sulphur by conc. \[HN{O_3}\]to a compound with the highest oxidation state of Sulphur is 288g.

Note: In this reaction nitric acid acts as an oxidizing agent which itself gets reduced and rhombic Sulphur is a reducing agent. The oxidation state varies from 0 to 6.