Answer:
[tex]\large \boxed{\text{93 g Mg(OH)}_{2}}[/tex]
Explanation:
We will need a balanced chemical equation with volumes, molar concentrations. and molar masses.
Mᵣ: 95.21
Mg(OH)₂ + 2HCl ⟶ MgCl₂ + 2H₂O
V/mL: 500 650
c/mol·L⁻¹: 3
1. Moles of HCl
[tex]\text{Moles of HCl}=\text{0.650 L HCl} \times \dfrac{\text{3 mol HCl}}{ \text{1 L HCl}} = \text{1.95 mol HCl}[/tex]
2. Moles of Mg(OH)₂
[tex]\text{Moles of MgOH}_{2} =\text{1.95 mol HCl}\times \dfrac{\text{1 mol Mg(OH)}_{2}}{\text{2 mol HCl}} =\text{0.975 mol MgOH}_{2}[/tex]
3. Mass of Mg(OH)₂
[tex]c = \text{0.975 mol Mg(OH)}_{2} \times \dfrac{\text{95.21 Mg(OH)}_{2}}{\text{1 mol Mg(OH)}_{2}} = \text{93 g Mg(OH)}_{2}\\\\\text{The reaction produces $\large \boxed{\textbf{93 g Mg(OH)}_{2}}$}[/tex]
