1) In this experiment, you will be mixing aqueous solutions of sodium carbonate and calcium chloride to produce solid calcium carbonate.

Question

Na2CO3 (aq) + CaCl2(aq) → 2 NaClaq) + CaCO3 (s)
Order the steps required to predict the volume (in mL) of 0.200 M calcium chloride needed to produce 2.00 g of calcium carbonate. There is an excess of sodium carbonate.

Answer 1

To predict the volume (in mL) of 0.200 M calcium chloride needed to produce 2.00 g of calcium carbonate, follow these steps:

1. **Calculate the molar mass of calcium carbonate (CaCO3):**
- Atomic masses: Calcium (Ca) = 40.08 g/mol, Carbon (C) = 12.01 g/mol, Oxygen (O) = 16.00 g/mol.
- Molar mass of CaCO3 = 40.08 + 12.01 + (3 x 16.00) = 100.09 g/mol.

2. **Determine the moles of calcium carbonate (CaCO3) you need to produce:**
- Mass of CaCO3 = 2.00 g.
- Moles of CaCO3 = mass / molar mass = 2.00 g / 100.09 g/mol ≈ 0.0200 mol.

3. **Use the stoichiometry of the reaction to find the moles of calcium chloride (CaCl2) needed:**
- According to the balanced equation, 1 mol of CaCl2 produces 1 mol of CaCO3.
- Therefore, moles of CaCl2 needed = moles of CaCO3 = 0.0200 mol.

4. **Calculate the volume of 0.200 M calcium chloride solution required:**
- Molarity (M) = moles of solute / liters of solution.
- Rearrange to find volume: volume (L) = moles of solute / Molarity = 0.0200 mol / 0.200 M = 0.100 L.

5. **Convert the volume from liters to milliliters:**
- 1 L = 1000 mL.
- Volume in mL = 0.100 L x 1000 mL/L = 100 mL.

So, the volume of 0.200 M calcium chloride needed to produce 2.00 g of calcium carbonate is 100 mL.