Consider the following reaction equation: \(2HCl + Ca(OH)_{2} \to CaCl_{2} + H_{2}O\). What is the volume of 0.1\(moldm^{-3}\) HCl that would completely neutralize 25\(cm^{3}\) of 0.3\(moldm^{-3}\) Ca(OH)\(_{2}\)?
-
150\(cm^{3}\)
- 75\(cm^{3}\)
- 30\(cm^{3}\)
- 25\(cm^{3}\)
The correct answer is: A
Explanation
Using the formula, \(\frac{C_{A}V_{A}}{C_{B}V_{B}} = \frac{N_{A}}{N_{B}}\), we have
\(\frac{0.1 \times V_{A}}{0.3 \times 25} = \frac{2}{1}\)
\(V_{A} = \frac{0.3 \times 25 \times 2}{0.1 \times 1}\)
\(V_{A} = 150cm^{3}\)