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 ____________ cm³ (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.]

Solution

Chemistry ma’am · 4 steps

Work backwards from the HCl: it tells you the leftover base, and the leftover equals what CO₂ used.

1. Leftover base. Moles of HCl = 0.1 × 0.040 = 0.004 mol. From Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O, each Ca(OH)₂ needs 2 HCl. So the unreacted Ca(OH)₂ is 0.002 mol.
2. Use the half. CO₂ used exactly half the initial Ca(OH)₂, so the other half was left over. The part that reacted with CO₂ is therefore also 0.002 mol.
3. Moles of CO2. From CO₂ + Ca(OH)₂ → CaCO₃ + H₂O, the ratio is 1 : 1. So n_(CO₂) = 0.002 mol.

   V = 0.002 × 22400 cm³ mol⁻¹ = 44.8 cm³

4. So. At 1 atm and 273 K one mole of gas occupies 22.4 L. So the container volume is 44.8 cm³, which rounds to 45.

The common slip: 90. Taking HCl and Ca(OH)₂ as 1 : 1 makes the excess 0.004 mol and gives about 90, not 45. One Ca(OH)₂ neutralises two HCl.

The idea to keep. In back-titration questions, find the excess reagent first from the titre, then subtract or use the given fraction. Always check the mole ratio of each reaction separately.

Answer from the NTA final answer key. Solution by PrepWiser.