Answer:
[tex]\boxed{\text{6.00 mol/L}}[/tex]
Explanation:
(a) Balanced equation
2HCl + Ca(OH)₂ ⟶ CaCl₂ + 2H₂O
(b) Moles of Ca(OH)₂
[tex]\text{Moles of base} = \text{2.00L} \times \dfrac{\text{1.50 mol}}{\text{1 L}} = \text{3.000 mol base}[/tex]
(c) Moles of HCl
[tex]\text{Moles of HCl} = \text{3.000 mol base} \times \dfrac{ \text{2 mol HCl}}{\text{1 mol base}} = \text{6.000 mol HCl}[/tex]
(d) Molar concentration of HCl
[tex]\text{Molar concentration} = \dfrac{\text{moles of solute}}{\text{litres of solution}}\\\\c = \dfrac{ n }{ V}\\\\c= \dfrac{ \text{6.000 mol}}{ \text{1.000 L}} = \text{6.00 mol/L}[/tex]
The molar concentration of the HCl was [tex]\boxed{\textbf{6.00 mol/L}}[/tex]
