Unit analysis for neutralization reaction
There is no special unit analysis.
It's a recommended practice to solve the problem algebraically first using proper notations for physical quantities, and plug the numeral values at the end minding the units — this way you reduce the chance of making the erroneous calculations and keep track of all units; as a bonus, you simplify handling significant figures.

The majority of problem solving in chemistry begins with writing down the balanced chemical reaction:

$$\ce{HCl + NaOH → NaCl + H2O}$$

So, to find the volume of hydrochloric acid $V(\ce{HCl})$, you may use its molar concentration $c(\ce{HCl})$ and the amount $n(\ce{HCl})$:

$$V(\ce{HCl}) = \frac{n(\ce{HCl})}{c(\ce{HCl})}$$

Unknown amount $n(\ce{HCl})$ can be found from the reaction's stoichiometry:

$$n(\ce{HCl}) = n(\ce{NaOH}) = \frac{m(\ce{NaOH})}{M(\ce{NaOH})}$$

where $m(\ce{NaOH})$ and $M(\ce{NaOH})$ are mass and molecular mass of sodium hydroxide, respectively.
Putting everything together and plugging in the appropriate numerical values, you arrive at the volume of approx. $\pu{100 mL}$:

$$
\begin{align}
V(\ce{HCl}) &amp;= \frac{m(\ce{NaOH})}{M(\ce{NaOH})\cdot c(\ce{HCl})} \\
  &amp;= \frac{\pu{2.02 g}}{\pu{40.00 g mol-1}\cdot \pu{0.500 mol L-1}} \\
  &amp;= \pu{0.101 L}
\end{align}
$$