What is the theoretical yield in grams of Ca(OH)2 if 2.04 g of KOH react with 3.02 g of CaCl2•2H2O as shown by the below equation?

CaCl2(aq) + 2KOH(aq) → Ca(OH)2(s) + 2KCl(aq)

In the part of the calculation that converts the number of grams to moles for CaCl2•2H2O, the two water molecules must be included in the molar mass needed to convert grams to moles (molar mass of CaCl2•2H2O is 147.0 g/mol).

The theoretical yield can be determined by finding the limiting reactant, which dictates the maximum amount of product (Ca(OH)₂) that can be formed. Here’s the process step by step:

### Given Information:
- \( \text{Mass of KOH} = 2.04 \, \text{g} \)
- \( \text{Mass of CaCl}_2 \cdot 2\text{H}_2\text{O} = 3.02 \, \text{g} \)
- Molar mass of \( \text{KOH} = 56.11 \, \text{g/mol} \)
- Molar mass of \( \text{CaCl}_2 \cdot 2\text{H}_2\text{O} = 147.0 \, \text{g/mol} \)
- Molar mass of \( \text{Ca(OH)}_2 = 74.09 \, \text{g/mol} \)

### Step 1: Convert grams of each reactant to moles

#### Moles of KOH:
\[
\text{Moles of KOH} = \frac{2.04 \, \text{g}}{56.11 \, \text{g/mol}} = 0.0364 \, \text{mol}
\]

#### Moles of CaCl₂·2H₂O:
\[
\text{Moles of CaCl}_2 \cdot 2\text{H}_2\text{O} = \frac{3.02 \, \text{g}}{147.0 \, \text{g/mol}} = 0.0205 \, \text{mol}
\]

### Step 2: Determine the limiting reactant

From the balanced chemical equation:
\[
\text{CaCl}_2 + 2\text{KOH} \rightarrow \text{Ca(OH)}_2 + 2\text{KCl}
\]

1 mole of CaCl₂ reacts with 2 moles of KOH to produce 1 mole of Ca(OH)₂.

- The amount of KOH required to react with 0.0205 moles of CaCl₂ is:
\[
\text{Required moles of KOH} = 0.0205 \, \text{mol CaCl}_2 \times 2 = 0.0410 \, \text{mol KOH}
\]
- However, we only have 0.0364 moles of KOH available, which is less than the required amount (0.0410 mol), meaning KOH is the limiting reactant.

### Step 3: Calculate the theoretical yield of Ca(OH)₂

Since KOH is the limiting reactant, we use its amount to determine how much Ca(OH)₂ is produced. From the equation, 2 moles of KOH produce 1 mole of Ca(OH)₂.

\[
\text{Moles of Ca(OH)}_2 = \frac{0.0364 \, \text{mol KOH}}{2} = 0.0182 \, \text{mol Ca(OH)}_2
\]

### Step 4: Convert moles of Ca(OH)₂ to grams

\[
\text{Mass of Ca(OH)}_2 = 0.0182 \, \text{mol} \times 74.09 \, \text{g/mol} = 1.35 \, \text{g}
\]

### Final Answer: 
The theoretical yield of \( \text{Ca(OH)}_2 \) is **1.35 grams**.

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Would you like more details on any part of the calculation?  
Here are some related questions for further exploration:

1. What is the actual yield, and how do you calculate percent yield?
2. How do you determine which reactant is the limiting reactant?
3. What happens if both reactants are provided in excess?
4. How does the molar mass of water affect hydrate calculations?
5. What are typical industrial uses of \( \text{Ca(OH)}_2 \)?

**Tip:** Always confirm that molar masses include all components of a compound, like water molecules in hydrates!

What is the theoretical yield in grams of Ca(OH)2 if 2.04 g of KOH react with 3.02 g of CaCl2•2H2O as shown by the below equation?

CaCl2(aq) + 2KOH(aq) → Ca(OH)2(s) + 2KCl(aq)

In the part of the calculation that converts the number of grams to moles for CaCl2•2H2O, the two water molecules must be included in the molar mass needed to convert grams to moles (molar mass of CaCl2•2H2O is 147.0 g/mol).

Learn how to calculate the theoretical yield of Ca(OH)2 when 2.04 g of KOH reacts with 3.02 g of CaCl2•2H2O. This example explores stoichiometry, mole-to-mass conversions, and determining the limiting reactant. Suitable for high school chemistry students.

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