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In this reaction: Mg(s) + 12 (s) → Mgl₂ (s), if 10.0 g of Mg reacts with 60.0 g of 12, and 64.00 g of Mgl, form, what is the percent yield?

In this reaction: Mg(s) + 12 (s) → Mgl₂ (s), if 10.0 g of Mg reacts with 60.0 g of 12, and 64.00 g of Mgl, form, what is the percent yield?

Chemistry
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 =
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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