$$SO_2Cl_2$$ on reaction with excess of water results into acidic mixture
$$SO_2Cl_2 + 2H_2O \rightarrow H_2SO_4 + 2HCl$$
16 moles of $$NaOH$$ is required for the complete neutralisation of the resultant acidic mixture. The number of moles of $$SO_2Cl_2$$ used is

The reaction of $$SO_2Cl_2$$ with excess water is:

$$SO_2Cl_2 + 2H_2O \rightarrow H_2SO_4 + 2HCl$$

So each mole of $$SO_2Cl_2$$ produces 1 mole of $$H_2SO_4$$ and 2 moles of $$HCl$$.

Neutralization reactions:

$$H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$$

$$HCl + NaOH \rightarrow NaCl + H_2O$$

For each mole of $$SO_2Cl_2$$:

- 1 mole $$H_2SO_4$$ requires 2 moles of $$NaOH$$

- 2 moles $$HCl$$ require 2 moles of $$NaOH$$

Total $$NaOH$$ required per mole of $$SO_2Cl_2 = 2 + 2 = 4$$ moles.

Given that 16 moles of $$NaOH$$ are required:

$$\text{Moles of } SO_2Cl_2 = \dfrac{16}{4} = 4$$

Therefore, the correct answer is Option C: $$4$$.