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:
- Molar mass of
- Molar mass of
- Molar mass of
Step 1: Convert grams of each reactant to moles
Moles of KOH:
Moles of CaCl₂·2H₂O:
Step 2: Determine the limiting reactant
From the balanced chemical equation:
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:
- 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.
Step 4: Convert moles of Ca(OH)₂ to grams
Final Answer:
The theoretical yield of is 1.35 grams.
Would you like more details on any part of the calculation?
Here are some related questions for further exploration:
- What is the actual yield, and how do you calculate percent yield?
- How do you determine which reactant is the limiting reactant?
- What happens if both reactants are provided in excess?
- How does the molar mass of water affect hydrate calculations?
- What are typical industrial uses of ?
Tip: Always confirm that molar masses include all components of a compound, like water molecules in hydrates!
Ask a new question for Free
By Image
Drop file here or Click Here to upload
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