Introductory Chemistry: A Foundation
9th Edition
ISBN:9781337399425
Author:Steven S. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Donald J. DeCoste
Chapter9: Chemical Quantities
Section: Chapter Questions
Problem 44QAP: Balance the following chemical equation, and then answer the question below....
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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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdddcaf98-1498-4874-b480-699455f18f4b%2F0eb3e1fb-d242-40b7-9772-42d4df39667f%2Fdean8vt_processed.png&w=3840&q=75)
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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