To solve this problem, we need to determine the oxidation state of cobalt ($$x$$) and the number of ammonia molecules ($$n$$) in the complex $$\mathrm{CoCl}_3 \cdot \mathrm{nNH}_3$$, given that it produces 2 moles of $$\mathrm{AgCl}$$ upon reaction with excess $$\mathrm{AgNO}_3$$.

First, let's write the reaction between the complex and $$\mathrm{AgNO}_3$$.

The precipitate formation indicates that some chloride ions are free (not coordinated to the metal). Since 2 moles of $$\mathrm{AgCl}$$ are formed, it indicates that there are 2 chloride ions that are free to react with $$\mathrm{AgNO}_3$$.

Therefore, we can conclude that in the complex, 1 chloride ion is coordinated to the cobalt, and 2 chloride ions are free. This can be represented as:

$$\mathrm{[Co(NH_3)_{n}Cl]Cl_2}$$

Now, let's determine the oxidation state of cobalt (Co). The charge on the entire complex should be zero, so we can write the charge balance equation as follows:

The sum of the charges is:

$$x + (0 \cdot n) + (-1 \cdot 1) + (-1 \cdot 2) = 0$$

Solving this, we get:

$$x - 1 - 2 = 0$$

$$x - 3 = 0$$

$$x = 3$$

So, the oxidation state of Co is 3.

Now, we need to find the value of $$n$$. Since the coordination number of Co in an octahedral complex is typically 6 and we have 1 chloride ion coordinated to Co, the number of ammonia molecules coordinated to Co would be:

$$n = 6 - 1 = 5$$

Finally, we calculate $$x + n$$:

$$x + n = 3 + 5 = 8$$

So, the value of "$$x + n$$" is 8.

Therefore, the correct option is:

Option D: 8