1. Write the Balanced Chemical Equation:

Na₂CO₃(aq) + CaCl₂•2H₂O(aq) → CaCO₃(s) + 2NaCl(aq) + 2H₂O(l)

2. Calculate the Molar Masses:

* Na₂CO₃: (2 * 22.99) + 12.01 + (3 * 16.00) = 105.99 g/mol

* CaCl₂•2H₂O: 40.08 + (2 * 35.45) + (2 * 18.02) = 147.02 g/mol

3. Convert Mass to Moles:

* Moles of Na₂CO₃: 1.00 g / 105.99 g/mol = 0.00943 mol

* Moles of CaCl₂•2H₂O: 1.00 g / 147.02 g/mol = 0.00680 mol

4. Determine the Limiting Reactant:

* Using the mole ratio from the balanced equation: The equation shows a 1:1 mole ratio between Na₂CO₃ and CaCl₂•2H₂O.

* Compare moles: Since we have fewer moles of CaCl₂•2H₂O (0.00680 mol) than Na₂CO₃ (0.00943 mol), CaCl₂•2H₂O is the limiting reactant.

Explanation:

The limiting reactant is the one that gets completely consumed first, preventing the reaction from proceeding further. In this case, even though we have more moles of Na₂CO₃, the reaction will stop when all of the CaCl₂•2H₂O is used up.