Unit analysis for neutralization reaction


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> 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? 
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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. $\pu{100 mL}$:


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


 2

The Unit/dimension analysis is in this case rather "using a cannon against sparrows".


The first spotted error is $\pu{500 g}$ of $\ce{HCl}$. Where did it come from ?


Note that the molar mass of $\ce{HCl}$ is about $\pu{36.5 g / mol}$.


1. Get the molar mass of $\ce{NaOH}$.
2. Calculate the amount of mols of $\ce{NaOH}$.
3. Calculate the equivalent amount of mols of $\ce{HCl}$.
4. Calculate the equivalent volume of $\pu{0.500 mol / l}$ $\ce{HCl}$.


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