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Consider the following reaction equation: \(2HCl + Ca(OH)_{2...

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 $cm^{3}$

View answer & explanation

Correct answer is A

Using the formula, $\frac{C_{A}V_{A}}{C_{B}V_{B}} = \frac{N_{A}}{N_{B}}$ , we have

$\frac{0.1 \times V_{A}}{0.3 \times 25} = \frac{2}{1}$

$V_{A} = \frac{0.3 \times 25 \times 2}{0.1 \times 1}$

$V_{A} = 150cm^{3}$

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