Unit analysis for neutralization reaction What volume of $\pu{0.500 mol L-1}$ $\ce{HCl}$ (in $\pu{mL}$) is required to neutralize $\pu{2.02 g}$ of solid $\ce{NaOH}$ pellets? 


So far I have tried regular unit analysis: 

2.02 g NaOH * 1mol NaOH/40g NaOH * 1mol HCl/1mol NaOH * 500g HCl/1mol HCl 
= 25.25 g HCl --> 25.5 mL HCl 

This answer doesn't make sense in the context of this neutralization reaction. Is there a type of unit analysis that is used specifically for determining the volume required for a 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. $\pu{100 mL}$:

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