Mole Concept and Stoichiometry — Question 5

(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