What volume of 0.132 M HCl is needed to neutralize 2.76 g of Mg(OH)_{2}?

**Reaction: 2 HCl + Mg(OH) _{2} → MgCl_{2} + H_{2}O**

Calculate moles HCl need to react with Mg(OH)_{2}:

molar mass Mg(OH)_{2} = 58.321 g/mol

$\mathbf{moles}\mathbf{}\mathbf{HCl}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{76}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{Mg}{\mathbf{\text{(OH)}}}_{\mathbf{2}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{Mg}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{2}}}}{\mathbf{58}\mathbf{.}\mathbf{321}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{Mg}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{2}}}}\mathbf{\times}\frac{\mathbf{2}\mathbf{}\mathbf{mol}\mathbf{}\mathbf{HCl}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{Mg}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{2}}}}$

**moles HCl = 0.09464 mol**

Calculate the volume of HCl solution needed:

What volume of 0.132 M HCl is needed to neutralize 2.76 g of Mg(OH)_{2}?

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