Question

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)

Answer 1

`"4.49 L"`.

Explanation:

Here's our balanced equation again:

`Ca(OH)_2 + 2HCl -> CaCl_2 + 2H_2O`

From this, we know that for every `1` mole of calcium hydroxide or `Ca(OH)_2` that reacts, `2` moles of `HCl` reacts.

So, our first step would be to find the number of moles of `Ca(OH)_2` that's in `"25 g"`.

To do this, we just need to divide `"25 g"` by the molar mass of `Ca(OH)_2`:
`25 / (Ca + 2 xx O + 2 xx H) = 25 / (40.08 + 2 xx 16.00 + 2 xx 1.008)`

`= 0.337` moles.

Therefore, because `2` moles of `HCl` react for every `1` mole of `Ca(OH)_2` that reacts, `0.337xx2 = 0.674` moles of `HCl` will react.

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