The chemical reaction between magnesium hydroxide and hydrochloric acid is given by: $$ \begin{aligned} \text{Mg(OH)}_2 + 2\text{HCl} &\rightarrow \text{MgCl}_2 + 2\text{H}_2\text{O} \end{aligned} $$ From the stoichiometry of the reaction, 1 mole of $\text{Mg(OH)}_2$ reacts with 2 moles of $\text{HCl}$. Therefore, the relationship between the moles is: $$ \begin{aligned} \frac{n_{\text{Mg(OH)}_2}}{1} &= \frac{n_{\text{HCl}}}{2} \end{aligned} $$ Given: - Mass of $\text{Mg(OH)}_2$, $w = 100 \text{ mg} = 0.1 \text{ g}$ - Molar mass of $\text{Mg(OH)}_2$, $M = 24.3 + 2 \times (16 + 1) = 58.3 \text{ g/mol}$ - Molar concentration of $\text{HCl}$, $S = 0.13 \text{ M}$ Substituting the values into the mole equation: $$ \begin{aligned} \frac{w}{M} &= \frac{S \times V}{2} \\ \frac{0.1}{58.3} &= \frac{0.13 \times V}{2} \\ \Rightarrow 0.001715 &= 0.065 \times V \\ \Rightarrow V &= \frac{0.001715}{0.065} \\ \Rightarrow V &\approx 0.02638 \text{ L} \end{aligned} $$ Thus, the volume of hydrochloric acid neutralized is approximately $0.02638 \text{ L}$.