Pure calcium carbonate and dilute hydrochloric acid are reacted and 2 litres of carbon dioxide was collected at 27°C and normal pressure.

CaCO₃ + 2HCl ⟶ CaCl₂ + H₂O + CO₂

Calculate:

(a) The mass of salt required.

(b) The mass of the acid required

(a) Given,

$\begin{matrix} \text{CaCO}$3 & + &2\text{HCl} & \longrightarrow & \text{CaCl}2 & + \text{H}2\text{O} + &\text{CO}2 \ 40 + 12 + 3(16) & & 2[1 + 35.5] & && & 1\text{ mole} \ = 40 + 12 + 48 && = 73 \text{ g} \ = 100 \text{ g} \ \end{matrix}

First convert the volume of carbon dioxide to STP:

V₁ = 2 L

T₁ = 27 + 273 K = 300 K

T₂ = 273 K

V₂ = ?

Using formula:

$\dfrac{\text{V}$1}{\text{T}1} = $\dfrac{\text{V}$2}{\text{T}2}

Substituting in the formula,

$\dfrac{2}{300}$ = $\dfrac{\text{V}_2}{273}$

V₂ = $\dfrac{2}{300}$ x 273 = 1.82 L

As, 22.4 L of carbon dioxide is obtained using 100 g CaCO₃

∴ 1.82 L of carbon dioxide is obtained from $\dfrac{100}{22.4}$ x 1.82

= 8.125 g of CaCO₃

(b) Similarly, 22.4 L of carbon dioxide is obtained using 73 g of acid

∴ 1.82 L of carbon dioxide is obtained from $\dfrac{73}{22.4}$ x 1.82

= 5.93 g of acid