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})?
-
A.
150(cm^{3}) -
B.
75(cm^{3}) -
C.
30(cm^{3}) -
D.
25(cm^{3})
Correct Answer: Option 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})