Recall:

$\overline{){\mathbf{Molarity}}{\mathbf{}}{\mathbf{\left(}}{\mathbf{M}}{\mathbf{\right)}}{\mathbf{=}}\frac{\mathbf{moles}\mathbf{}\mathbf{solute}\mathbf{}\mathbf{\left(}\mathbf{mol}\mathbf{\right)}}{\mathbf{volume}\mathbf{}\mathbf{solution}\mathbf{}\mathbf{\left(}\mathbf{L}\mathbf{\right)}}}$

Determine moles HCl first based on the stoichiometry of the reaction:

NaCl(*s*) + H_{2}SO_{4}(*l*) ⟶ HCl(*g*) + NaHSO_{4}(*s*)

$\mathbf{moles}\mathbf{}\mathbf{HCl}\mathbf{=}\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{NaCl}}\left(\frac{1\overline{)\mathrm{mol}\mathrm{NaCl}}}{58.44g\overline{)\mathrm{NaCl}}}\right)\left(\frac{1\mathrm{mol}\mathrm{HCl}}{1\overline{)\mathrm{mol}\mathrm{NaCl}}}\right)\mathbf{=}$**0.428 mol**

What volume of 0.750 M hydrochloric acid solution can be prepared from the HCl produced by the reaction of 25.0 g of NaCl with excess sulfuric acid?

NaCl(*s*) + H_{2}SO_{4}(*l*) ⟶ HCl(*g*) + NaHSO_{4}(*s*)

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