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A 1.00 L volume of [tex]$HCl[tex]$[/tex] reacted completely with 2.00 L of [tex]$[/tex]1.50 M Ca(OH)_2$[/tex] according to the balanced chemical equation below:

[tex]\[ 2 HCl + Ca(OH)_2 \rightarrow CaCl_2 + 2 H_2O \][/tex]

What was the molarity of the [tex]$HCl$[/tex] solution?

A. 0.375 M
B. 1.50 M
C. 3.00 M
D. 6.00 M


Sagot :

To determine the molarity of the HCl solution, follow these steps systematically:

### Step 1: Identify the Given Information
- Volume of HCl solution: \(1.00 \, \text{L}\)
- Volume of Ca(OH)₂ solution: \(2.00 \, \text{L}\)
- Molarity of Ca(OH)₂ solution: \(1.50 \, \text{M}\)

### Step 2: Write the Balanced Chemical Equation
The balanced chemical equation for the reaction is:
[tex]\[ 2 \, \text{HCl} + \text{Ca(OH)}_2 \rightarrow \text{CaCl}_2 + 2 \, \text{H}_2\text{O} \][/tex]

### Step 3: Calculate Moles of Ca(OH)₂
Using the volume and molarity of the Ca(OH)₂ solution, calculate the number of moles of Ca(OH)₂:
[tex]\[ \text{Moles of Ca(OH)}_2 = \text{Molarity} \times \text{Volume} \][/tex]
[tex]\[ \text{Moles of Ca(OH)}_2 = 1.50 \, \text{M} \times 2.00 \, \text{L} = 3.00 \, \text{moles} \][/tex]

### Step 4: Determine Moles of HCl Required
From the balanced chemical equation, we see that 2 moles of HCl react with 1 mole of Ca(OH)₂. Therefore, the moles of HCl needed are twice the moles of Ca(OH)₂:
[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]

### Step 5: Calculate the Molarity of HCl Solution
To find the molarity of the HCl solution, we use the formula:
[tex]\[ \text{Molarity} = \frac{\text{Moles of Solute}}{\text{Volume of Solution}} \][/tex]
[tex]\[ \text{Molarity of HCl solution} = \frac{6.00 \, \text{moles}}{1.00 \, \text{L}} = 6.00 \, \text{M} \][/tex]

### Conclusion
The molarity of the HCl solution is \(6.00 \, \text{M}\). Therefore, the correct answer is:
[tex]\[ \boxed{6.00 \, \text{M}} \][/tex]