Question

For the given reaction:
CaCO3 + 2HCl → CaCl2 + H2O + CO2

If 90 g CaCO3 is added to 300 mL of HCl which contains 38.55% HCl by mass and has density 1.13 g mL-1,
then which of the following option is correct?

Given molar mass of H, Cl, Ca and O are 1, 35.5, 40 and 16 g mol-1 respectively.

• A. 60.32 g of HCl remains unreacted
• B. 32.85 g of CaCO\(_3\) remains unreacted
• C. 97.30 g of HCl reacted
• D. 64.97 g of HCl remains unreacted

Hint:

Always check limiting reagent by comparing mole ratios. Excess reagent remains unreacted, while limiting reagent decides product yield.

Answer 1

The Correct Option is B

Step 1: Calculate moles of CaCO3.
Molar mass of CaCO3 = 40 + 12 + (3 × 16) = 100 g mol-1
Moles of CaCO3 = 90 / 100 = 0.9 mol

Step 2: Calculate mass of HCl solution.
Mass of solution = 300 × 1.13 = 339 g
Mass of HCl = 0.3855 × 339 ≈ 130.8 g

Step 3: Calculate moles of HCl.
Molar mass of HCl = 1 + 35.5 = 36.5 g mol-1
Moles of HCl = 130.8 / 36.5 ≈ 3.58 mol

Step 4: Limiting reagent check.
Reaction: CaCO3 + 2HCl → CaCl2 + H2O + CO2
0.9 mol CaCO3 requires 2 × 0.9 = 1.8 mol HCl
Available HCl = 3.58 mol (excess)
Hence, CaCO3 is the limiting reagent.

Step 5: Conclusion.
Since HCl is in excess, all CaCO3 reacts completely.
No CaCO3 remains unreacted.

Final Conclusion:
CaCO3 is the limiting reagent and reacts completely.
Hence, the option stating CaCO3 as the limiting reagent is correct.