Some CO2 gas was kept in a sealed container at a pressure of 1 atm and at 273 K. This entire amount of CO2 gas was later passed through an aqueous solution of Ca(OH)2. The excess unreacted Ca(OH)2 was later neutralized with 0.1 M of 40 mL HCl. If the volume of the sealed container of CO2 was x, then x is ______ cm 3 (nearest integer). [Given : The entire amount of CO2(g) reacted with exactly half the initial amount of Ca(OH)2 present in the aqueous solution.]

1. Stoichiometry of the two reactions

1. CO₂ with limewater
   $\text{Ca(OH)}_2 + \text{CO}_2 \longrightarrow \text{CaCO}_3 + \text{H}_2\text{O}$
   • 1 mole CO₂ needs 1 mole Ca(OH)₂.

2. Excess limewater with HCl
   $\text{Ca(OH)}_2 + 2\,\text{HCl} \longrightarrow \text{CaCl}_2 + 2\,\text{H}_2\text{O}$
   • 1 mole Ca(OH)₂ is neutralised by 2 moles HCl.

2. Interpreting the statement in the question

The sentence “The entire amount of CO₂ reacted with exactly half the initial amount of Ca(OH)₂” means:

• If the starting number of moles of Ca(OH)₂ is $n_i$, then: • moles that reacted with CO₂ $= \frac{n_i}{2}$
  • moles left over $= n_i - \frac{n_i}{2} = \frac{n_i}{2}$

3. Using the titration data to find $n_i$

The leftover Ca(OH)₂ was titrated with 40 mL of 0.1 M HCl.

Number of moles of HCl used:

$\text{moles HCl} = 0.1\,\text{mol L}^{-1} \times 0.040\,\text{L} = 0.004\,\text{mol}$

Because 2 HCl neutralise 1 Ca(OH)₂:

$\text{moles Ca(OH)}_2\,\text{left} = \frac{0.004}{2} = 0.002\,\text{mol}$

But leftover equals $\frac{n_i}{2}$, so:

$\frac{n_i}{2} = 0.002 \;\Rightarrow\; n_i = 0.004\,\text{mol}$

4. Moles of CO₂ that reacted

Moles of Ca(OH)₂ that reacted with CO₂: $\frac{n_i}{2} = 0.002\,\text{mol}$
By the 1 : 1 stoichiometry, moles of CO₂ $= 0.002\,\text{mol}$.

5. Converting moles of CO₂ to volume at 273 K & 1 atm

At these conditions, 1 mole of an ideal gas occupies 22.4 L (22 400 cm³).

$V = n \times 22.4\,\text{L} = 0.002 \times 22.4 = 0.0448\,\text{L}$

Convert to cubic centimetres (cm³):

$0.0448\,\text{L} \times 1000 = 44.8\,\text{cm}^3$

Nearest integer ⇒ 45 cm³.

🌟 What is happening?

Imagine you have a balloon full of the same stuff that makes soda fizzy (CO₂).
You squeeze all of that gas into a jar of limewater (that’s just water with a chalk-making powder called Ca(OH)₂).
The CO₂ and the limewater react and make chalk (CaCO₃), but only half of the limewater actually reacts.
Whatever limewater is left over is then washed away (neutralised) with some stomach-acid-like liquid called HCl.

🏁 What do we need?

We want to know how big the balloon (container) of CO₂ was.
To do that, we must:

1. Count how much limewater stayed unreacted using the acid it took to finish it off.
2. Work out how much CO₂ reacted (it equals the limewater that reacted).
3. Use the gas law at cold-weather conditions (273 K and 1 atm) to change moles of gas into volume.

That will tell us the size of the balloon!

Step-by-Step Solution

1. Calculate moles of HCl used in the titration
   $n_{\text{HCl}} = 0.1\,\text{M} \times 0.040\,\text{L} = 0.004\,\text{mol}$

2. Convert HCl moles to leftover Ca(OH)₂ moles
   $\text{Ca(OH)}_2 + 2\,\text{HCl} \rightarrow \text{CaCl}_2 + 2\,\text{H}_2\text{O}$
   $n_{\text{Ca(OH)}_2\,\text{left}} = \frac{n_{\text{HCl}}}{2} = \frac{0.004}{2} = 0.002\,\text{mol}$

3. Relate leftover moles to initial moles
   Since leftover equals half the initial amount: $\frac{n_i}{2} = 0.002 \;\Rightarrow\; n_i = 0.004\,\text{mol}$

4. Find moles of Ca(OH)₂ that reacted with CO₂
   $n_{\text{reacted}} = \frac{n_i}{2} = 0.002\,\text{mol}$

5. Determine moles of CO₂ (1 : 1 ratio)
   $n_{\text{CO}_2} = 0.002\,\text{mol}$

6. Convert moles of CO₂ to volume at 1 atm & 273 K
   $V = nRT/P = n \times 22.4\,\text{L} = 0.002 \times 22.4 = 0.0448\,\text{L}$
   $V = 44.8\,\text{cm}^3$

7. Round to nearest integer
   $\boxed{45\,\text{cm}^3}$