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

To solve this problem, we need to determine the amount of hydrochloric acid (HCl) that remains unreacted after the chemical reaction described. Let's analyze the situation step-by-step:

1. Determine the moles of calcium carbonate (\(\mathrm{CaCO_3}\)):
   The molar mass of \(\mathrm{CaCO_3}\) is calculated as: 40 + 12 + 3 \times 16 = 100 \, \text{g/mol}. Given mass of \(\mathrm{CaCO_3}\) is \(90 \, \text{g}\). So, the moles of \(\mathrm{CaCO_3}\) are: \frac{90}{100} = 0.9 \, \text{mol}.
2. Calculate the amount of HCl in the solution:
   • Mass of HCl in the solution:
     The solution has \(38.55\%\) HCl by weight. Therefore, in \(300 \, \text{mL}\) of solution, the mass of solution is: 300 \times 1.13 = 339 \, \text{g}.
   • Mass of HCl: \frac{38.55}{100} \times 339 = 130.78 \, \text{g}.
3. Determine the theoretical amount of HCl required to react with \(\mathrm{CaCO_3}\):
   The balanced chemical reaction is:
   \[ \mathrm{CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2} \] According to the stoichiometry, 2 moles of HCl are needed per mole of \(\mathrm{CaCO_3}\). Thus, the moles of HCl needed are: 2 \times 0.9 = 1.8 \, \text{mol}. The molar mass of HCl is \(36.5 \, \text{g/mol}\), so the mass needed is: 1.8 \times 36.5 = 65.7 \, \text{g}.
4. Calculate the remaining unreacted HCl:
   • The initial mass of HCl was \(130.78 \, \text{g}\).
   • The mass of HCl that reacted is \(65.7 \, \text{g}\).
   • Hence, the unreacted mass of HCl is: 130.78 - 65.7 = 65.08 \, \text{g} (approx \(64.97 \, \text{g}\)).

Thus, the correct answer is \(64.97 \, \text{g}\) of HCl remains unreacted. This matches the given correct option in the problem statement.