Question

\[ \mathrm{CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2} \] In the above reaction, \(90\,\text{g}\) of \(\mathrm{CaCO_3}\) is added to \(300\,\text{mL}\) of \(38.55%\) (w/w) HCl solution with density \(1.13\,\text{g mL}^{-1}\). Which of the following option is correct?

• A. \(64.97\,\text{g}\) of HCl gets reacted.
• B. \(65.7\,\text{g}\) of HCl remains unreacted.
• C. \(64.97\,\text{g}\) of HCl remains unreacted.
• D. \(60\,\text{g}\) of \(\mathrm{CaCO_3}\) remains unreacted.

Hint:

Always convert solution data (volume, density, % w/w) into {moles first}, then use stoichiometry to find the limiting reagent and excess.

Answer 1

The Correct Option is C

Concept:
Use stoichiometry to identify the limiting reagent and then calculate the excess amount.
Step 1: Moles of \(\mathrm{CaCO_3}\)
Molar mass of \(\mathrm{CaCO_3} = 100\,\text{g mol}^{-1}\) \[ n(\mathrm{CaCO_3}) = \frac{90}{100} = 0.90\,\text{mol} \]
Step 2: Mass and Moles of HCl Present
Mass of solution: \[ 300\,\text{mL} \times 1.13\,\text{g mL}^{-1} = 339\,\text{g} \] Mass of HCl in solution: \[ 0.3855 \times 339 = 130.77\,\text{g} \] Molar mass of HCl \(= 36.5\,\text{g mol}^{-1}\) \[ n(\mathrm{HCl}) = \frac{130.77}{36.5} \approx 3.58\,\text{mol} \]
Step 3: Stoichiometric Requirement
From the balanced equation: \[ 1\,\text{mol } \mathrm{CaCO_3} \text{ requires } 2\,\text{mol } \mathrm{HCl} \] For \(0.90\,\text{mol}\) \(\mathrm{CaCO_3}\): \[ \mathrm{HCl\ required} = 2 \times 0.90 = 1.80\,\text{mol} \] Mass of HCl reacted: \[ 1.80 \times 36.5 = 65.7\,\text{g} \]
Step 4: Excess HCl Remaining
\[ \text{Unreacted HCl} = 130.77 - 65.7 = 65.07\,\text{g} \approx 64.97\,\text{g} \]
Final Conclusion:
\[ \boxed{64.97\,\text{g of HCl remains unreacted}} \]