# How many liters of 0.150 M HCl would be required to react completely with 25.00 grams of calcium hydroxide?

## Considering the following equation Ca(OH)2(s) +2HCl(aq) ---> CaCl2(aq) +2H2O(l)

Mar 23, 2018

$\text{4.49 L}$.

#### Explanation:

Here's our balanced equation again:

$C a {\left(O H\right)}_{2} + 2 H C l \to C a C {l}_{2} + 2 {H}_{2} O$

From this, we know that for every $1$ mole of calcium hydroxide or $C a {\left(O H\right)}_{2}$ that reacts, $2$ moles of $H C l$ reacts.

So, our first step would be to find the number of moles of $C a {\left(O H\right)}_{2}$ that's in $\text{25 g}$.

To do this, we just need to divide $\text{25 g}$ by the molar mass of $C a {\left(O H\right)}_{2}$:
$\frac{25}{C a + 2 \times O + 2 \times H} = \frac{25}{40.08 + 2 \times 16.00 + 2 \times 1.008}$

$= 0.337$ moles.

Therefore, because $2$ moles of $H C l$ react for every $1$ mole of $C a {\left(O H\right)}_{2}$ that reacts, $0.337 \times 2 = 0.674$ moles of $H C l$ will react.

The final step is to calculate volume in $L$, given $\text{0.150 M}$, or $\text{0.150 mol/L}$, and $0.674$ moles.
Knowing this, the volume would be: $\text{0.674 mol"/"0.150 mol/L" = "4.49 L}$.