Question

The volume of CO\(_2\) liberated in litres at STP when 25 g of CaCO\(_3\) is treated with dilute HCl containing 14.6 g of HCl is:

• A. 22.4 L
• B. 11.2 L
• C. 4.48 L
• D. 44.8 L

Hint:

At STP, 1 mole of any ideal gas occupies 22.4 L.

Answer 1

The Correct Option is C

Step 1: Balanced Chemical Equation \[ \ch{CaCO3 + 2 HCl - > CaCl2 + H2O + CO2} \] Step 2: Calculate Moles of Reactants
\begin{itemize} \item Moles of \ch{CaCO3}: \[ \text{Molar mass of \ch{CaCO3}} = 40 + 12 + 3 \times 16 = 100\ \text{g/mol} \] \[ \text{Moles of \ch{CaCO3}} = \frac{25\ \text{g}}{100\ \text{g/mol}} = 0.25\ \text{moles} \] \item Moles of \ch{HCl}: \[ \text{Molar mass of \ch{HCl}} = 1 + 35.5 = 36.5\ \text{g/mol} \] \[ \text{Moles of \ch{HCl}} = \frac{14.6\ \text{g}}{36.5\ \text{g/mol}} = 0.4\ \text{moles} \] \end{itemize} Step 3: Determine Limiting Reagent
From the balanced equation: \[ 1\ \text{mole \ch{CaCO3}} \text{ reacts with } 2\ \text{moles \ch{HCl}} \] \[ \text{Required \ch{HCl} for 0.25 moles \ch{CaCO3}} = 0.25 \times 2 = 0.5\ \text{moles} \] Available \ch{HCl} is only 0.4 moles, so \ch{HCl is the limiting reagent}. Step 4: Calculate Moles of \ch{CO2 Produced}
From the stoichiometry: \[ 2\ \text{moles \ch{HCl}} \text{ produce } 1\ \text{mole \ch{CO2}} \] \[ \text{Moles of \ch{CO2}} = \frac{0.4}{2} = 0.2\ \text{moles} \] Step 5: Volume of \ch{CO2 at STP}
At STP (Standard Temperature and Pressure): \[ 1\ \text{mole of gas} = 22.4\ \text{L} \] \[ \text{Volume of \ch{CO2}} = 0.2 \times 22.4 = 4.48\ \text{L} \] Final Answer \[ \boxed{4.48\ \text{L}} \]