Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![### Chemical Reaction and Percent Yield Calculation
#### Problem Statement:
In this reaction:
\[ \text{Mg (s)} + \text{I}_2 \text{(s)} \rightarrow \text{MgI}_2 \text{(s)} \]
If 10.0 g of Mg reacts with 60.0 g of I\(_2\), and 64.00 g of MgI\(_2\) forms, what is the percent yield?
### Explanation:
In this problem, we are given the masses of reactants and the mass of the product formed. The goal is to determine the percent yield of the reaction.
#### Steps to Solve:
1. **Calculate the Theoretical Yield:**
- Find the molar masses:
- Mg = 24.305 g/mol
- I\(_2\) = 2 \times 126.90 g/mol = 253.80 g/mol
- MgI\(_2\) = 24.305 g/mol + 2 \times 126.90 g/mol = 278.105 g/mol
- Determine the limiting reagent by calculating the moles of each reactant:
- Moles of Mg: \( \frac{10.0 \text{ g}}{24.305 \text{ g/mol}} \approx 0.411 \text{ mol} \)
- Moles of I\(_2\): \( \frac{60.0 \text{ g}}{253.80 \text{ g/mol}} \approx 0.236 \text{ mol} \)
- The stoichiometry of the reaction tells us that 1 mol of Mg reacts with 1 mol of I\(_2\). Therefore, I\(_2\) is the limiting reagent.
- Calculate the theoretical yield of MgI\(_2\) (using limiting reagent I\(_2\)):
- \(0.236 \text{ mol of I}_2 \times 278.105 \text{ g/mol MgI}_2 \approx 65.67 \text{ g of MgI}_2\)
2. **Calculate the Percent Yield:**
- Percent Yield = \(\frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%\)
- Actual yield = 64.00 g (given)
- Percent Yield =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F18d0a7dc-afe1-42e7-9349-84d2cfc69c53%2F5f2c755b-84a3-4a74-83a6-d20dae18da06%2F3zjqlzg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Chemical Reaction and Percent Yield Calculation
#### Problem Statement:
In this reaction:
\[ \text{Mg (s)} + \text{I}_2 \text{(s)} \rightarrow \text{MgI}_2 \text{(s)} \]
If 10.0 g of Mg reacts with 60.0 g of I\(_2\), and 64.00 g of MgI\(_2\) forms, what is the percent yield?
### Explanation:
In this problem, we are given the masses of reactants and the mass of the product formed. The goal is to determine the percent yield of the reaction.
#### Steps to Solve:
1. **Calculate the Theoretical Yield:**
- Find the molar masses:
- Mg = 24.305 g/mol
- I\(_2\) = 2 \times 126.90 g/mol = 253.80 g/mol
- MgI\(_2\) = 24.305 g/mol + 2 \times 126.90 g/mol = 278.105 g/mol
- Determine the limiting reagent by calculating the moles of each reactant:
- Moles of Mg: \( \frac{10.0 \text{ g}}{24.305 \text{ g/mol}} \approx 0.411 \text{ mol} \)
- Moles of I\(_2\): \( \frac{60.0 \text{ g}}{253.80 \text{ g/mol}} \approx 0.236 \text{ mol} \)
- The stoichiometry of the reaction tells us that 1 mol of Mg reacts with 1 mol of I\(_2\). Therefore, I\(_2\) is the limiting reagent.
- Calculate the theoretical yield of MgI\(_2\) (using limiting reagent I\(_2\)):
- \(0.236 \text{ mol of I}_2 \times 278.105 \text{ g/mol MgI}_2 \approx 65.67 \text{ g of MgI}_2\)
2. **Calculate the Percent Yield:**
- Percent Yield = \(\frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%\)
- Actual yield = 64.00 g (given)
- Percent Yield =
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