https://file.aiquickdraw.com/mathbot/file/1727851406619ra0hoyha.jpg

The problem describes a reaction between calcium hydroxide (Ca(OH)₂) and hydrochloric acid (HCl), and you are asked to determine the limiting reagent (pembatas), the excess reagent (sisa), and the moles of calcium chloride (CaCl₂) produced.

**Reaction equation:**
\[ \text{Ca(OH)}_2 + 2 \text{HCl} \rightarrow \text{CaCl}_2 + 2 \text{H}_2\text{O} \]

Let's break it down:

### Given Information:
- Volume of Ca(OH)₂ solution = 100 mL = 0.1 L
- Molarity of Ca(OH)₂ solution = 0.1 M
- Volume of HCl solution = 100 mL = 0.1 L
- Molarity of HCl solution = 0.1 M

### Step 1: Calculate moles of each reactant

- Moles of Ca(OH)₂:
\[
\text{Moles of Ca(OH)}_2 = M \times V = 0.1 \, \text{M} \times 0.1 \, \text{L} = 0.01 \, \text{mol}
\]

- Moles of HCl:
\[
\text{Moles of HCl} = M \times V = 0.1 \, \text{M} \times 0.1 \, \text{L} = 0.01 \, \text{mol}
\]

### Step 2: Analyze the stoichiometry of the reaction
From the balanced chemical equation:
\[
1 \, \text{mol of Ca(OH)}_2 \text{ reacts with } 2 \, \text{mol of HCl}
\]
So, 0.01 mol of Ca(OH)₂ will require:
\[
0.01 \, \text{mol of Ca(OH)}_2 \times 2 = 0.02 \, \text{mol of HCl}
\]

### Step 3: Determine the limiting reagent
- You only have 0.01 mol of HCl, but the reaction requires 0.02 mol of HCl to completely react with 0.01 mol of Ca(OH)₂. Therefore, HCl is the **limiting reagent** (pembatas).

### Step 4: Calculate the amount of CaCl₂ produced
From the equation:
\[
1 \, \text{mol of Ca(OH)}_2 \text{ produces } 1 \, \text{mol of CaCl}_2
\]
Since the limiting reagent (HCl) limits the reaction, we base our calculation on the moles of HCl:
\[
\frac{0.01}{2} = 0.005 \, \text{mol of CaCl}_2
\]

### Step 5: Identify the excess reagent
Ca(OH)₂ is in excess. To find how much remains, we calculate how much reacted:
- Moles of Ca(OH)₂ that reacted = moles of HCl used ÷ 2:
\[
\frac{0.01}{2} = 0.005 \, \text{mol}
\]
- Moles of Ca(OH)₂ left = 0.01 mol (initial) - 0.005 mol (reacted) = **0.005 mol remains**.

### Final Answers:
- **Limiting reagent (pembatas):** HCl
- **Excess reagent (sisa):** Ca(OH)₂ (0.005 mol remaining)
- **Moles of CaCl₂ produced:** 0.005 mol

Would you like further details or clarification?

Here are 5 related questions to expand on this topic:
1. How do you calculate the mass of CaCl₂ produced from the given moles?
2. What is the significance of identifying the limiting reagent in a chemical reaction?
3. How can the concentration of the remaining Ca(OH)₂ be calculated?
4. What happens if the volumes of HCl and Ca(OH)₂ are reversed in quantity?
5. How would the reaction change if the molarity of HCl was doubled?

**Tip:** Always check the stoichiometric ratios from the balanced equation before determining the limiting reagent.

100 mL of 0.1 M Ca(OH)₂ is reacted with 100 mL of 0.1 M HCl based on the reaction: Ca(OH)₂ + 2 HCl → CaCl₂ + 2 H₂O. Determine the limiting reagent, excess reagent, and moles of CaCl₂ produced.

Learn how to determine the limiting reagent, excess reagent, and moles of CaCl₂ produced when 100 mL of 0.1 M Ca(OH)₂ reacts with 100 mL of 0.1 M HCl. Understand stoichiometric calculations and reactant analysis.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation