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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 60.36 g of Mgl₂ 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 60.36 g of Mgl₂ 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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### Calculating Percent Yield in a Chemical Reaction **Problem Statement:** In this reaction: \[ \text{Mg} (s) + \text{I}_2 (s) \rightarrow \text{MgI}_2 (s) \] If 10.0 g of Mg reacts with 60.0 g of I<sub>2</sub>, and 60.36 g of MgI<sub>2</sub> forms, what is the percent yield? **Details of Calculation:** 1. **Balanced Chemical Equation**: \[ \text{Mg} (s) + \text{I}_2 (s) \rightarrow \text{MgI}_2 (s) \] 2. **Molar Mass Calculation**: - Molar mass of \( \text{Mg} \): 24.305 g/mol - Molar mass of \( \text{I}_2 \): 2 × 126.904 g/mol = 253.808 g/mol - Molar mass of \( \text{MgI}_2 \): 24.305 g/mol + 2 × 126.904 g/mol = 278.113 g/mol 3. **Stoichiometric Calculations**: - Moles of \( \text{Mg} \): \( \frac{10.0 \, \text{g}}{24.305 \, \text{g/mol}} = 0.4112 \, \text{mol} \) - Moles of \( \text{I}_2 \): \( \frac{60.0 \, \text{g}}{253.808 \, \text{g/mol}} = 0.2364 \, \text{mol} \) Since Mg and I2 react in a 1:1 ratio: - \( \text{Mg} \) is the limiting reagent since its molar quantity is less than that of \( \text{I}_2 \). 4. **Theoretical Yield**: - Moles of \( \text{MgI}_2 \) produced: 0.4112 mol (from \( \text{Mg} \)) - Theoretical mass of \( \text{MgI}_2 \): \( 0.4112 \, \text{mol} \times 278.113 \, \text{g
Transcribed Image Text:### Calculating Percent Yield in a Chemical Reaction **Problem Statement:** In this reaction: \[ \text{Mg} (s) + \text{I}_2 (s) \rightarrow \text{MgI}_2 (s) \] If 10.0 g of Mg reacts with 60.0 g of I<sub>2</sub>, and 60.36 g of MgI<sub>2</sub> forms, what is the percent yield? **Details of Calculation:** 1. **Balanced Chemical Equation**: \[ \text{Mg} (s) + \text{I}_2 (s) \rightarrow \text{MgI}_2 (s) \] 2. **Molar Mass Calculation**: - Molar mass of \( \text{Mg} \): 24.305 g/mol - Molar mass of \( \text{I}_2 \): 2 × 126.904 g/mol = 253.808 g/mol - Molar mass of \( \text{MgI}_2 \): 24.305 g/mol + 2 × 126.904 g/mol = 278.113 g/mol 3. **Stoichiometric Calculations**: - Moles of \( \text{Mg} \): \( \frac{10.0 \, \text{g}}{24.305 \, \text{g/mol}} = 0.4112 \, \text{mol} \) - Moles of \( \text{I}_2 \): \( \frac{60.0 \, \text{g}}{253.808 \, \text{g/mol}} = 0.2364 \, \text{mol} \) Since Mg and I2 react in a 1:1 ratio: - \( \text{Mg} \) is the limiting reagent since its molar quantity is less than that of \( \text{I}_2 \). 4. **Theoretical Yield**: - Moles of \( \text{MgI}_2 \) produced: 0.4112 mol (from \( \text{Mg} \)) - Theoretical mass of \( \text{MgI}_2 \): \( 0.4112 \, \text{mol} \times 278.113 \, \text{g
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