Let's analyze the reaction provided in the image and solve the problem step by step.

Given:

• 100 mL of 0.1 M Ca(OH)₂ (Calcium Hydroxide)
• 100 mL of 0.1 M HCl (Hydrochloric Acid)
• Reaction:
  $\text{Ca(OH)}_2 + 2 \text{HCl} \rightarrow \text{CaCl}_2 + 2 \text{H}_2\text{O}$

Steps to solve:

1. Determine the number of moles of each reactant:

   • Ca(OH)₂: $\text{moles of Ca(OH)}_2 = M \times V = 0.1 \, \text{M} \times 0.1 \, \text{L} = 0.01 \, \text{mol}$

   • HCl: $\text{moles of HCl} = M \times V = 0.1 \, \text{M} \times 0.1 \, \text{L} = 0.01 \, \text{mol}$

2. Determine the limiting reagent: The balanced equation shows that 1 mole of Ca(OH)₂ reacts with 2 moles of HCl.
   $\text{Ca(OH)}_2 + 2 \text{HCl} \rightarrow \text{CaCl}_2 + 2 \text{H}_2\text{O}$ We need 2 moles of HCl for every mole of Ca(OH)₂. But since we only have 0.01 mol of HCl, which is not enough to fully react with the 0.01 mol of Ca(OH)₂, HCl is the limiting reagent.

3. Determine the amount of product (CaCl₂) formed: According to the stoichiometry of the reaction, 1 mole of Ca(OH)₂ produces 1 mole of CaCl₂.
   Therefore, the amount of CaCl₂ produced is based on the limiting reagent HCl: $\text{moles of CaCl}_2 = \frac{1}{2} \times 0.01 \, \text{mol of HCl} = 0.005 \, \text{mol}$

4. Determine the excess reactant (Ca(OH)₂) left: Since 0.01 mol of HCl reacts with 0.005 mol of Ca(OH)₂ (based on the stoichiometric ratio), the leftover Ca(OH)₂ is: $0.01 \, \text{mol} - 0.005 \, \text{mol} = 0.005 \, \text{mol}$

Final answers:

• Limiting reagent: HCl
• Excess reagent: Ca(OH)₂ (0.005 mol left)
• Moles of CaCl₂ produced: 0.005 mol

Let me know if you need further clarifications or a more detailed explanation!

Here are some follow-up questions:

1. How does stoichiometry help in determining the limiting reagent?
2. What would happen if the volume of HCl were doubled in this reaction?
3. How is the concept of molarity used in this problem?
4. What is the role of the balanced chemical equation in predicting reaction outcomes?
5. Could this reaction be done with a different base, such as NaOH? How would the results change?

Tip:

Always balance the chemical equation before starting stoichiometry problems, as it ensures accurate calculations of reactants and products.