Gas released in the following reaction will be: $2Na+2HN{O}_3\to 2NaN{O}_3+?$

We are aware that sodium is a metal and that nitric acid is a potent oxidizer and acid. You must be able to recollect the action of an acid on a metal in order to respond to this question. Hydrocarbons that are unsaturated are reduced using the gas that was emitted.

A metal and an acid react by a straightforward displacement reaction. Because it is an acid and releases a proton, nitric acid acts as an oxidising agent in this situation. Because nitric acid is monobasic, each molecule of the acid only emits one hydrogen ion. The metal takes the place of the hydrogen ion that the acid released.

Sodium metal, which is the metal used in the process above, has a potent capacity to transfer electrons and transform into a positive ion. The metal forms a salt with the conjugate base of the acid because it is a good oxidising agent and has a tendency to accept the electrons and establish a bond with the metal. In contrast, the hydrogen ion absorbs the electron produced by the metal.
Two hydrogen atoms are created when two molecules of the metal react with two molecules of the acid, and as a result, hydrogen gas is emitted.

The reaction occurring can be given as:

$\mathrm{Na}+{\mathrm{HNO}}_3\left(dil\right)\to {\mathrm{NaNO}}_3+{\mathrm{H}}_2\uparrow$

Hence, the correct option is B.