When 8 g of CaCO3 are combined with 4 g of HCl, which substance is the limiting reagent in the reaction CaCO3 + 2 HCl → CaCl2 + CO2 + H2O?

• A. HCl
• B. CaCO3
• C. Both are limiting
• D. Neither is limiting

Answer: (A)

Verified step by step guidance

1. Write the balanced chemical equation: \(\text{CaCO}_3 + 2\ \text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O}\).
2. Calculate the number of moles of each reactant using their molar masses. For CaCO\(_3\), molar mass is approximately 100 g/mol, so moles of CaCO\(_3\) = \(\frac{8\ \text{g}}{100\ \text{g/mol}}\). For HCl, molar mass is approximately 36.5 g/mol, so moles of HCl = \(\frac{4\ \text{g}}{36.5\ \text{g/mol}}\).
3. Determine the mole ratio required by the balanced equation: 1 mole of CaCO\(_3\) reacts with 2 moles of HCl. Use this ratio to find the amount of HCl needed to completely react with the given moles of CaCO\(_3\).
4. Compare the actual moles of HCl available to the moles required to react with CaCO\(_3\). If the available moles of HCl are less than required, HCl is the limiting reagent; otherwise, CaCO\(_3\) is limiting.
5. Conclude which reactant is limiting based on the comparison, since the limiting reagent is the one that will be completely consumed first, limiting the amount of product formed.