Sure, let's solve this step-by-step.
We start by recalling the balanced chemical equation provided:
[tex]\[ 2 \text{HCl} + \text{Ca(OH)}_2 \rightarrow \text{CaCl}_2 + 2 \text{H}_2\text{O} \][/tex]
From the balanced equation, we see that 2 moles of HCl react with 1 mole of [tex]\(\text{Ca(OH)}_2\)[/tex].
1. Determine the moles of [tex]\(\text{Ca(OH)}_2\)[/tex]:
We know the volume and the molarity of [tex]\(\text{Ca(OH)}_2\)[/tex]:
- Volume of [tex]\(\text{Ca(OH)}_2\)[/tex]: 2.00 L
- Molarity of [tex]\(\text{Ca(OH)}_2\)[/tex]: 1.50 M
The number of moles of [tex]\(\text{Ca(OH)}_2\)[/tex] can be calculated as:
[tex]\[
\text{Moles of } \text{Ca(OH)}_2 = \text{Molarity} \times \text{Volume} = 1.50 \, \text{M} \times 2.00 \, \text{L} = 3.00 \, \text{moles}
\][/tex]
2. Use the stoichiometry to find the moles of HCl:
According to the balanced equation, 2 moles of HCl react with 1 mole of [tex]\(\text{Ca(OH)}_2\)[/tex]. Therefore, for the 3.00 moles of [tex]\(\text{Ca(OH)}_2\)[/tex] we have, we need twice that amount of HCl.
[tex]\[
\text{Moles of HCl} = 2 \times \text{Moles of } \text{Ca(OH)}_2 = 2 \times 3.00 = 6.00 \, \text{moles}
\][/tex]
3. Calculate the molarity of HCl:
We know the volume of the HCl solution is 1.00 L. Molarity is defined as the number of moles of solute divided by the volume of the solution in liters.
[tex]\[
\text{Molarity of HCl} = \frac{\text{Moles of HCl}}{\text{Volume of HCl}} = \frac{6.00 \, \text{moles}}{1.00 \, \text{L}} = 6.00 \, \text{M}
\][/tex]
Therefore, the molarity of the HCl solution is [tex]\( 6.00 \, \text{M} \)[/tex].
So the correct answer is:
- 6.00 M
