1. Write the balanced chemical equation:

CaCO₃(s) + 2 HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)

2. Calculate the molar mass of CaCO₃:

* Ca: 40.08 g/mol

* C: 12.01 g/mol

* O: 16.00 g/mol (x3)

* Molar mass of CaCO₃ = 40.08 + 12.01 + (16.00 x 3) = 100.09 g/mol

3. Convert grams of CaCO₃ to moles:

* Moles of CaCO₃ = (0.055 g) / (100.09 g/mol) = 0.00055 mol

4. Use the stoichiometry of the balanced equation to find moles of HCl:

* From the balanced equation, 1 mole of CaCO₃ reacts with 2 moles of HCl.

* Moles of HCl = (0.00055 mol CaCO₃) * (2 mol HCl / 1 mol CaCO₃) = 0.0011 mol HCl

Therefore, 0.055 g of calcium carbonate can neutralize 0.0011 moles of HCl.