Question

a 1.00 l volume of hcl reacted completely with 2.00 l of 1.50 m ca(oh)2 according to the balanced chemical equation below.
2hcl + ca(oh)2 → cacl2 + 2h2o
what was the molarity of the hcl solution?
∘ 0.375 m
∘ 1.50 m
∘ 3.00 m
∘ 6.00 m

Explanation:

Step1: Calculate moles of \( Ca(OH)_2 \)

Molarity formula: \( M = \frac{n}{V} \), so \( n = M \times V \).
For \( Ca(OH)_2 \), \( M = 1.50 \, M \), \( V = 2.00 \, L \).
\( n_{Ca(OH)_2} = 1.50 \, \frac{mol}{L} \times 2.00 \, L = 3.00 \, mol \).

Step2: Use stoichiometry to find moles of \( HCl \)

From balanced equation \( 2HCl + Ca(OH)_2
ightarrow CaCl_2 + 2H_2O \), the mole ratio \( \frac{n_{HCl}}{n_{Ca(OH)_2}} = \frac{2}{1} \).
So \( n_{HCl} = 2 \times n_{Ca(OH)_2} = 2 \times 3.00 \, mol = 6.00 \, mol \).

Step3: Calculate molarity of \( HCl \)

Volume of \( HCl \) is \( 1.00 \, L \). Molarity \( M = \frac{n}{V} \).
\( M_{HCl} = \frac{6.00 \, mol}{1.00 \, L} = 6.00 \, M \).

Answer:

6.00 M