The amount of calcium carbonate that reacts with 500cc of 0.5N hydrochloric acid can be determined using stoichiometry and the concept of molarity.

1. Understanding the problem:
- We are given the volume of hydrochloric acid (HCl) solution, which is 500cc or 500 mL.
- The concentration of the hydrochloric acid solution is given as 0.5N, which means it is 0.5 normal.
- We need to find the amount of calcium carbonate (CaCO3) that reacts with the given hydrochloric acid solution.

2. Balancing the chemical equation:
- The balanced chemical equation for the reaction between calcium carbonate and hydrochloric acid is:
CaCO3 + 2HCl -> CaCl2 + H2O + CO2

3. Understanding molarity:
- The concentration of a solution is given in terms of molarity (M).
- Molarity is defined as the number of moles of solute per liter of solution.
- In this case, the concentration of the hydrochloric acid solution is given as 0.5N, which means it contains 0.5 moles of HCl per liter of solution.

4. Calculating the number of moles of HCl:
- To determine the amount of calcium carbonate that reacts, we first need to calculate the number of moles of hydrochloric acid.
- We know the volume of the hydrochloric acid solution is 500cc or 500 mL.
- We can convert this volume to liters by dividing it by 1000: 500/1000 = 0.5 L.
- Now, we can calculate the number of moles of HCl using the formula: moles = concentration (M) x volume (L).
- Substituting the values, moles of HCl = 0.5N x 0.5 L = 0.25 moles.

5. Determining the stoichiometry:
- According to the balanced chemical equation, one mole of calcium carbonate reacts with two moles of hydrochloric acid.
- This means that for every mole of HCl, 0.5 moles of calcium carbonate will react.

6. Calculating the amount of calcium carbonate:
- We already know the number of moles of HCl is 0.25.
- Using the stoichiometry, we can calculate the number of moles of calcium carbonate that reacts.
- Moles of CaCO3 = 0.25 moles of HCl x 0.5 moles of CaCO3/1 mole of HCl = 0.125 moles.

7. Converting moles to grams:
- Finally, we can convert the moles of calcium carbonate to grams using its molar mass.
- The molar mass of calcium carbonate (CaCO3) is 100.09 g/mol.
- Mass of CaCO3 = 0.125 moles x 100.09 g/mol = 12.5125 grams.

Therefore, the amount of calcium carbonate that reacts with 500cc of 0.5N hydrochloric acid is approximately 12.5125 grams.