# What mass of hydrogen chloride is necessary to neutralize a 75.1*g mass of calcium hydroxide?

##### 1 Answer
Feb 28, 2017

Approx. $74 \cdot g$..........

#### Explanation:

We need (i) a stoichiometric equation:

$C a {\left(O H\right)}_{2} \left(s\right) + 2 H C l \left(a q\right) \rightarrow C a C {l}_{2} \left(a q\right) + 2 {H}_{2} O \left(l\right)$.

And (ii), equivalent quantities of calcium hydroxide:

$\text{Moles of calcium hydroxide} = \frac{75.1 \cdot g}{74.09 \cdot g \cdot m o {l}^{-} 1} = 1.01 \cdot m o l$.

And for an equivalent quantity of hydrochloric acid, we thus need $2 \times 1.01 \cdot m o l \times 36.46 \cdot g \cdot m o {l}^{-} 1 = 73.6 \cdot g$.

Why did I double the molar quantity of calcium hydroxide?

Given that we would normally use $\text{34% conc. hydrochloric acid}$, which has an approx. molarity of $10.9 \cdot m o l \cdot {L}^{-} 1$, what is the volume of hydrochloric acid required for equivalence?