Calculate the volume (in mL) of 0.100 M CaCl2 needed to produce 1.00 g of . There is an excess of Na₂CO3. Na₂CO3(aq) +CaCl₂(aq) - → > 2 NaCl(aq) + CaCO3(s) Molar mass of calcium carbonate = 100.09 g/mol Volume of calcium chloride = mL CaCO3(s)

Chemistry
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**Chemical Reaction and Calculation**

**Reaction Equation:**
\[
\text{Na}_2\text{CO}_3(aq) + \text{CaCl}_2(aq) \rightarrow 2 \text{NaCl}(aq) + \text{CaCO}_3(s)
\]

**Objective:**
Calculate the volume (in mL) of 0.100 M \(\text{CaCl}_2\) needed to produce 1.00 g of \(\text{CaCO}_3(s)\).

**Given Information:**

- **Molar Mass of Calcium Carbonate (\(\text{CaCO}_3\))**: 100.09 g/mol
- **There is an excess of \(\text{Na}_2\text{CO}_3\).**

**Calculation:**

To find the required volume of calcium chloride solution, calculate the moles of \(\text{CaCO}_3\) produced, then use stoichiometry and molarity concepts to find the volume needed.

- **Formula for Moles Calculation**: 
  \[
  \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}
  \]

- Calculate moles of \(\text{CaCO}_3\):
  \[
  \text{moles of CaCO}_3 = \frac{1.00 \, \text{g}}{100.09 \, \text{g/mol}}
  \]

- Use the balanced chemical equation to determine moles of \(\text{CaCl}_2\) needed:
  - 1 mole of \(\text{CaCl}_2\) produces 1 mole of \(\text{CaCO}_3\).

- Use the molarity equation to calculate the volume:
  \[
  \text{Volume (L)} = \frac{\text{moles of }\text{CaCl}_2}{\text{Molarity (M)}}
  \]

Convert volume from liters to milliliters.

**Volume of calcium chloride = [Answer in mL]**

**Diagram Explanation:**

There are no diagrams or graphs in the text.
Transcribed Image Text:**Chemical Reaction and Calculation** **Reaction Equation:** \[ \text{Na}_2\text{CO}_3(aq) + \text{CaCl}_2(aq) \rightarrow 2 \text{NaCl}(aq) + \text{CaCO}_3(s) \] **Objective:** Calculate the volume (in mL) of 0.100 M \(\text{CaCl}_2\) needed to produce 1.00 g of \(\text{CaCO}_3(s)\). **Given Information:** - **Molar Mass of Calcium Carbonate (\(\text{CaCO}_3\))**: 100.09 g/mol - **There is an excess of \(\text{Na}_2\text{CO}_3\).** **Calculation:** To find the required volume of calcium chloride solution, calculate the moles of \(\text{CaCO}_3\) produced, then use stoichiometry and molarity concepts to find the volume needed. - **Formula for Moles Calculation**: \[ \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \] - Calculate moles of \(\text{CaCO}_3\): \[ \text{moles of CaCO}_3 = \frac{1.00 \, \text{g}}{100.09 \, \text{g/mol}} \] - Use the balanced chemical equation to determine moles of \(\text{CaCl}_2\) needed: - 1 mole of \(\text{CaCl}_2\) produces 1 mole of \(\text{CaCO}_3\). - Use the molarity equation to calculate the volume: \[ \text{Volume (L)} = \frac{\text{moles of }\text{CaCl}_2}{\text{Molarity (M)}} \] Convert volume from liters to milliliters. **Volume of calcium chloride = [Answer in mL]** **Diagram Explanation:** There are no diagrams or graphs in the text.
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