To determine the molarity of the HCl solution, let's follow a step-by-step approach:
1. Identify the volume and molarity of Ca(OH)2:
- Volume of Ca(OH)2 solution: 2.00 liters
- Molarity of Ca(OH)2 solution: 1.50 M (mol/L)
2. Calculate the moles of Ca(OH)2:
[tex]\[
\text{Moles of Ca(OH)2} = \text{Volume of Ca(OH)2} \times \text{Molarity of Ca(OH)2}
\][/tex]
[tex]\[
\text{Moles of Ca(OH)2} = 2.00 \, \text{L} \times 1.50 \, \text{mol/L} = 3.00 \, \text{moles}
\][/tex]
3. Use the balanced chemical equation to find the mole ratio between Ca(OH)2 and HCl:
[tex]\[
2 \, \text{HCl} + \text{Ca(OH)2} \rightarrow \text{CaCl2} + 2 \, \text{H2O}
\][/tex]
According to the equation, 1 mole of Ca(OH)2 reacts with 2 moles of HCl.
4. Calculate the moles of HCl required:
[tex]\[
\text{Moles of HCl} = 2 \times \text{Moles of Ca(OH)2}
\][/tex]
[tex]\[
\text{Moles of HCl} = 2 \times 3.00 \, \text{moles} = 6.00 \, \text{moles}
\][/tex]
5. Identify the volume of the HCl solution:
- Volume of HCl solution: 1.00 liter
6. Calculate the molarity of the HCl solution:
[tex]\[
\text{Molarity of HCl} = \frac{\text{Moles of HCl}}{\text{Volume of HCl}}
\][/tex]
[tex]\[
\text{Molarity of HCl} = \frac{6.00 \, \text{moles}}{1.00 \, \text{L}} = 6.00 \, \text{M}
\][/tex]
Therefore, the molarity of the HCl solution is 6.00 M. The correct answer is:
6.00 M.
