In a stoichiometry lab, you react 2.00 mol of HCl with 1.00 mol of CaCO3. According to the balanced equation CaCO3 + 2 HCl → CaCl2 + CO2 + H2O, what is the limiting reactant?

• A. CO2
• B. HCl
• C. CaCO3
• D. CaCl2

Answer: (C)

Verified step by step guidance

1. Write down the balanced chemical equation: \(\mathrm{CaCO_3 + 2\ HCl \rightarrow CaCl_2 + CO_2 + H_2O}\).
2. Identify the mole ratio between the reactants from the balanced equation: 1 mole of \(\mathrm{CaCO_3}\) reacts with 2 moles of \(\mathrm{HCl}\).
3. Compare the given moles of each reactant to the mole ratio: You have 1.00 mol of \(\mathrm{CaCO_3}\) and 2.00 mol of \(\mathrm{HCl}\).
4. Calculate how much \(\mathrm{HCl}\) is required to completely react with 1.00 mol of \(\mathrm{CaCO_3}\) using the mole ratio: \$1.00\ \mathrm{mol\ CaCO_3} \times \frac{2\ \mathrm{mol\ HCl}}{1\ \mathrm{mol\ CaCO_3}} = 2.00\ \mathrm{mol\ HCl}$.
5. Since the available moles of \(\mathrm{HCl}\) exactly match the required amount to react with all of the \(\mathrm{CaCO_3}\), neither reactant is in excess, but if any less \(\mathrm{HCl}\) were present, it would limit the reaction. Here, \(\mathrm{CaCO_3}\) is the limiting reactant because it will be completely consumed first when reacting with the given \(\mathrm{HCl}\).