2 L of 0.2 M $$H_2SO_4$$ is reacted with 2 L of 0.1 M NaOH solution, the molarity of the resulting product $$Na_2SO_4$$ in the solution is _____ millimolar.

We have 2 L of 0.2 M $$H_2SO_4$$ and 2 L of 0.1 M NaOH solution being mixed together. We need to find the molarity of $$Na_2SO_4$$ formed in the resulting solution.

First, let us calculate the moles of each reactant.

Moles of $$H_2SO_4 = 0.2 \times 2 = 0.4$$ mol

Moles of NaOH $$= 0.1 \times 2 = 0.2$$ mol

The reaction between $$H_2SO_4$$ and NaOH is:

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

From the stoichiometry, 1 mol of $$H_2SO_4$$ requires 2 mol of NaOH. So 0.2 mol of NaOH would react with $$\frac{0.2}{2} = 0.1$$ mol of $$H_2SO_4$$.

Since we have 0.4 mol of $$H_2SO_4$$ but only 0.2 mol of NaOH, the NaOH is the limiting reagent. All 0.2 mol of NaOH reacts with 0.1 mol of $$H_2SO_4$$, and the remaining 0.3 mol of $$H_2SO_4$$ stays unreacted.

From the stoichiometry, 0.1 mol of $$H_2SO_4$$ produces 0.1 mol of $$Na_2SO_4$$.

Now, the total volume of the resulting solution is $$2 + 2 = 4$$ L.

The molarity of $$Na_2SO_4$$ is:

$$\text{Molarity} = \frac{0.1}{4} = 0.025 \text{ M} = 25 \text{ millimolar}$$

Hence, the correct answer is 25.