Mole Concept
The limiting reagent in the reaction 2Al + 3Cl2 → 2AlCl3 when 10.0 g Al (M=27) reacts with 20.0 g Cl2 (M=71) is:
Select the correct option:
Solution
Cl2
- Balanced Equation: 2 Al+3 Cl2→2 AlCl3.
- Calculate Initial Moles:
- nAl=2710.0≈0.370 mol.
- nCl2=7120.0≈0.282 mol.
- Determine Stoichiometric Requirement:
- According to the reaction, 1 mol of Al requires 1.5 mol of Cl2
- For 0.370 mol Al, required Cl2=0.370×23=0.555 mol.
- Identify Limiting Reagent:
- Since the required Cl2 (0.555 mol) is greater than the available Cl2 (0.282 mol),
- Cl2 is the limiting reagent.
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About This Question
- Subject
- chemistry
- Chapter
- some basic concepts in chemistry
- Topic
- mole concept
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
Cl2
- Balanced Equation: 2 Al+3 Cl2→2 AlCl3.
- Calculate Initial Moles:
- nAl=2710.0≈0.370 mol.
- nCl2=7120.0≈0.282 mol.
- Determine Stoichiometric Requirement:
- According to the reaction, 1 mol of Al requires 1.5 mol of Cl2
- For 0.370 mol Al, required Cl2=0.370×23=0.555 mol.
- Identify Limiting Reagent:
- Since the required Cl2 (0.555 mol) is greater than the available Cl2 (0.282 mol),
- Cl2 is the limiting reagent.
This medium difficulty chemistry question is from the chapter some basic concepts in chemistry, covering the topic of mole concept. It appeared in the 2025 exam.
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