How do I calculate the grams of a product after a reaction? If you mix $50\,\mathrm{mL}$ of $2.5\,\mathrm{M}$ $\ce{HCl}$ with $100\,\mathrm{mL}$ $2.5\,\mathrm{M}$ $\ce{NaOH}$, you will get sodium chloride. If such reaction occurs only at $50~\%$, how many grams of sodium chloride will you get?


Well, I guess that the reaction is like

$$\ce{HCl + NaOH \to NaCl + H2O}$$

Since the concentration is the same for both reagents ($2.5\,\mathrm{M}$), I imagine that the concentration will be the same for the products?

So we got $2.5\,\mathrm{M}$ of $\ce{NaCl}$. However, to get the grams I first need to know the moles of $\ce{NaCl}$. I know that the concentration is $2.5\,\mathrm{M}$, but what of it? What can I do to get the moles of the resulting $\ce{NaCl}$?
The concentration $c$ of a solution is the molar quantity of the solute $n$ devided by the volume of the solution $V$:

$\begin{equation}
c = \frac{n}{V}
\end{equation}$

as you state, for example for HCl 2.5 mol/L (or, equivalent, 2.5 M).  Inversely, if you know volume and concentration of your solution, you may access the molar quantity of your solute.  Actually, there is such an equation for each of the solutions, indepent from each other; you may index the variables accordingly, too.

Yet watch, there is a catch: mixing the two solutions together increases the total volume, and simultaneously decreases the concentration of the specis in question -- without altering the molar quantity of $\ce{Na+}$, nor $\ce{Cl-}$ ions present.  Hence "2.5 M of NaCl" (as measure of concentration) in the solution in common is incorrect.

Because of the initially asked question, one of the solutions contains the limiting reagent.  Figure out which one -- this includes the stoichiometry in the equation of chemical reaction. Knowing the molecular amount $n$ of it, your are able to determine the amount of sodium chloride formed; initially as moles, later (via molecular mass) in grams.