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what is the theoretical and percent yield of the reaction?

what is the theoretical and percent yield of the reaction?

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 Yield Calculation #### Reaction Overview In the following reaction: \[ \text{Al} + \text{Br}_2 \rightarrow \text{AlBr}_3 \] If 14.5 g of Al and 46.6 g of Br₂ are reacted, and 26 g of AlBr₃ is produced, what is the theoretical and percent yield of the reaction? #### Given Data - Mass of Al: 14.5 g - Mass of Br₂: 46.6 g - Mass of AlBr₃ produced: 26 g #### Calculation Steps 1. **Determine the Molar Mass of Reactants and Products:** - Molar mass of Al (Aluminum): 26.98 g/mol - Molar mass of Br₂ (Bromine): 79.90 g/mol * 2 = 159.80 g/mol - Molar mass of AlBr₃ (Aluminum Bromide): 26.98 g/mol + (79.90 g/mol * 3) = 266.68 g/mol 2. **Convert Masses to Moles:** - Moles of Al: \(\frac{14.5 \text{ g}}{26.98 \text{ g/mol}} \approx 0.537 \text{ mol}\) - Moles of Br₂: \(\frac{46.6 \text{ g}}{159.80 \text{ g/mol}} \approx 0.292 \text{ mol}\) 3. **Identify the Limiting Reactant:** - The balanced equation: 2Al + 3Br₂ → 2AlBr₃ - According to the stoichiometry, 2 moles of Al react with 3 moles of Br₂. - Moles of Al needed for 0.292 mol of Br₂: \(\frac{2}{3} \times 0.292 \text{ mol} = 0.195 \text{ mol}\) - Since there are more moles of Al than needed, Br₂ is the limiting reactant. 4. **Calculate Theoretical Yield:** - Moles of AlBr₃ produced by 0.292 mol of Br₂: As per the balanced equation, 3 moles of Br₂ produce 2 moles of Al
Transcribed Image Text:### Chemical Reaction Yield Calculation #### Reaction Overview In the following reaction: \[ \text{Al} + \text{Br}_2 \rightarrow \text{AlBr}_3 \] If 14.5 g of Al and 46.6 g of Br₂ are reacted, and 26 g of AlBr₃ is produced, what is the theoretical and percent yield of the reaction? #### Given Data - Mass of Al: 14.5 g - Mass of Br₂: 46.6 g - Mass of AlBr₃ produced: 26 g #### Calculation Steps 1. **Determine the Molar Mass of Reactants and Products:** - Molar mass of Al (Aluminum): 26.98 g/mol - Molar mass of Br₂ (Bromine): 79.90 g/mol * 2 = 159.80 g/mol - Molar mass of AlBr₃ (Aluminum Bromide): 26.98 g/mol + (79.90 g/mol * 3) = 266.68 g/mol 2. **Convert Masses to Moles:** - Moles of Al: \(\frac{14.5 \text{ g}}{26.98 \text{ g/mol}} \approx 0.537 \text{ mol}\) - Moles of Br₂: \(\frac{46.6 \text{ g}}{159.80 \text{ g/mol}} \approx 0.292 \text{ mol}\) 3. **Identify the Limiting Reactant:** - The balanced equation: 2Al + 3Br₂ → 2AlBr₃ - According to the stoichiometry, 2 moles of Al react with 3 moles of Br₂. - Moles of Al needed for 0.292 mol of Br₂: \(\frac{2}{3} \times 0.292 \text{ mol} = 0.195 \text{ mol}\) - Since there are more moles of Al than needed, Br₂ is the limiting reactant. 4. **Calculate Theoretical Yield:** - Moles of AlBr₃ produced by 0.292 mol of Br₂: As per the balanced equation, 3 moles of Br₂ produce 2 moles of Al
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