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

[tex]
2 \, \text{HCl} + \text{Ca(OH)}_2 \rightarrow \text{CaCl}_2 + 2 \, \text{H}_2\text{O}
[/tex]

What was the molarity of the HCl solution?

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



Answer :

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