Question

What property of $Conc.{H_2}S{O_4}$ is made use of in each of the following cases? Give an equation for the reaction in each case.
A. In the production of $HCl$ gas when it reacts with a chloride.
B. In the production of $CO$ from $HCOOH$.
C. As a source of hydrogen by diluting it and adding a strip of magnesium.
D. In the preparation of Sulphur dioxide by warming a mixture of conc. Sulphuric acid and copper turnings.
E. Hydrogen sulphide gas is passed through concentrated Sulphuric acid.

Answer 1

Acids are the chemical compounds that can easily release a proton, these compounds are very acidic and have lower values of pH. Concentrated acids are more acidic and concentrated Sulphuric acid is an acid. Though, it is an acid it behaves as oxidizing or reducing agent depending upon the other reagents in chemical reaction.

Complete answer:
When concentrated Sulphuric acid is treated with sodium chloride due to its reducing property, it behaves as a volatile acid.
$NaCl + Conc.{H_2}S{O_4} \to NaHS{O_4} + HCl$
When concentrated Sulphuric acid is treated formic acid, it behaves as a dehydrating agent.
$HCOOH\xrightarrow{{Conc.{H_2}S{O_4}}}CO + {H_2}O$
When magnesium is added to concentrated Sulphuric acid liberation of hydrogen gas takes place, as hydrogen is more reactive than magnesium.
$Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}$
Due to its oxidizing nature, conc. Sulphuric acid when added to copper turnings Sulphur dioxide is liberated.
$Cu + {H_2}S{O_4} \to CuS{O_4} + 2{H_2}O + S{O_2}$
Due to its oxidizing nature, conc. Sulphuric acid when added to hydrogen sulphide gas, Sulphur dioxide is liberated.
${H_2}S + {H_2}S{O_4} \to S + 2{H_2}O + S{O_2}$

Note:
Sometimes, concentrated Sulphuric acid behaves as oxidizer and sometimes as reducer depending upon the other reagent in the chemical reaction. The liberation of Sulphur dioxide gas takes place due to its oxidizing nature.