Math Problem Statement

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).

Solution

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:

  • Mass of KOH=2.04g\text{Mass of KOH} = 2.04 \, \text{g}
  • Mass of CaCl22H2O=3.02g\text{Mass of CaCl}_2 \cdot 2\text{H}_2\text{O} = 3.02 \, \text{g}
  • Molar mass of KOH=56.11g/mol\text{KOH} = 56.11 \, \text{g/mol}
  • Molar mass of CaCl22H2O=147.0g/mol\text{CaCl}_2 \cdot 2\text{H}_2\text{O} = 147.0 \, \text{g/mol}
  • Molar mass of Ca(OH)2=74.09g/mol\text{Ca(OH)}_2 = 74.09 \, \text{g/mol}

Step 1: Convert grams of each reactant to moles

Moles of KOH:

Moles of KOH=2.04g56.11g/mol=0.0364mol\text{Moles of KOH} = \frac{2.04 \, \text{g}}{56.11 \, \text{g/mol}} = 0.0364 \, \text{mol}

Moles of CaCl₂·2H₂O:

Moles of CaCl22H2O=3.02g147.0g/mol=0.0205mol\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: CaCl2+2KOHCa(OH)2+2KCl\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: Required moles of KOH=0.0205mol CaCl2×2=0.0410mol KOH\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.

Moles of Ca(OH)2=0.0364mol KOH2=0.0182mol 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

Mass of Ca(OH)2=0.0182mol×74.09g/mol=1.35g\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 Ca(OH)2\text{Ca(OH)}_2 is 1.35 grams.


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 Ca(OH)2\text{Ca(OH)}_2?

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

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Math Problem Analysis

Mathematical Concepts

Stoichiometry
Mole-to-Mass Conversion
Limiting Reactant

Formulas

Moles of substance = mass / molar mass
Mole ratio from balanced chemical equation
Mass of product = moles × molar mass

Theorems

Law of Conservation of Mass

Suitable Grade Level

Grades 10-12

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