The gas produced by treating an aqueous solution of ammonium chloride with sodium nitrite is

When an aqueous solution of ammonium chloride ($$NH_4Cl$$) is treated with sodium nitrite ($$NaNO_2$$), the following reaction takes place:

$$NH_4Cl + NaNO_2 \rightarrow NaCl + N_2 + 2H_2O$$

This is a well-known reaction used in the laboratory for the preparation of pure nitrogen gas ($$N_2$$). The ammonium ion ($$NH_4^+$$) acts as a mild reducing agent and the nitrite ion ($$NO_2^-$$) acts as a mild oxidizing agent. The nitrogen in $$NH_4^+$$ (oxidation state -3) and the nitrogen in $$NO_2^-$$ (oxidation state +3) combine to form $$N_2$$ (oxidation state 0). This is a comproportionation (disproportionation) reaction where nitrogen in two different oxidation states combines to form nitrogen in an intermediate oxidation state.

The gas produced is $$N_2$$, which matches Option B.

The answer is $$\boxed{\text{Option B: } N_2}$$.