Addition of $$H_2SO_4$$ to $$BaO_2$$ produces

We need to determine the products when $$H_2SO_4$$ is added to $$BaO_2$$ (barium peroxide). $$BaO_2$$ is barium peroxide, which contains the peroxide ion $$O_2^{2-}$$.

When barium peroxide reacts with dilute sulfuric acid, it undergoes a double displacement reaction:

$$BaO_2 + H_2SO_4 \rightarrow BaSO_4 + H_2O_2$$

The $$Ba^{2+}$$ ion combines with $$SO_4^{2-}$$ to form the insoluble salt $$BaSO_4$$. The peroxide ion $$O_2^{2-}$$ combines with $$2H^+$$ to form $$H_2O_2$$ (hydrogen peroxide). This is actually one of the historical methods for producing hydrogen peroxide.

The correct answer is Option D.