Chapter 1: Problem 40
Aluminum metal reacts with bromine, a red-brown liquid with a noxious odor. The reaction 1.40 is vigorous and produces aluminum bromide, a white crystalline substance. A sample of \(27.0 \mathrm{~g}\) of aluminum yields \(266.7 \mathrm{~g}\) of aluminum bromide. How many grams of bromine react with \(15.0 \mathrm{~g}\) of aluminum?
Short Answer
Step by step solution
Understand the Reaction
Find Molar Masses
Calculate Moles of Aluminum Used
Relate Moles of Aluminum Used
Calculate Mass of Bromine Needed for 27.0 g Aluminum
Scale the Reaction for 15.0 g Aluminum
Conclusion
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chemical Reactions
- The reactants in this case are aluminum and bromine.
- The product is aluminum bromide.
Moles and Molar Mass
- Molar mass is the mass of one mole of a substance, measured in grams per mole (g/mol). This value varies from element to element and compound to compound.
- For example, the molar mass of aluminum is about 26.98 g/mol, while that of bromine (as a diatomic molecule, Br\(_2\)) is approximately 159.80 g/mol.
Balanced Chemical Equations
- In the given reaction, the balanced equation is: \ \(2 \text{Al} + 3 \text{Br}_2 \rightarrow 2 \text{AlBr}_3\)
- This indicates that two moles of aluminum react with three moles of bromine to produce two moles of aluminum bromide.