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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Transcribed Image Text:**Question 11 of 14**
**Determine the number of atoms of O in 57.4 moles of Al₂(CO₃)₃.**
This problem involves calculating the number of oxygen atoms in a given amount of moles of a chemical compound. The compound in question is Aluminum Carbonate, Al₂(CO₃)₃, which contains carbon and oxygen in its carbonate groups.
### Explanation
- **Molecular Formula**: Al₂(CO₃)₃
- Each molecule of Aluminum Carbonate consists of 2 aluminum atoms, 3 carbonate ions.
- Each carbonate ion, CO₃²⁻, contains 1 carbon atom and 3 oxygen atoms.
- **Calculating Oxygen Atoms**:
- In each formula unit of Al₂(CO₃)₃, there are 3 carbonate ions.
- Therefore, there are \(3 \times 3 = 9\) oxygen atoms in each molecule of Al₂(CO₃)₃.
- Given 57.4 moles of Al₂(CO₃)₃, the total number of moles of oxygen atoms = 57.4 moles × 9.
- The number of atoms is found using Avogadro's number (\(6.022 \times 10^{23}\) atoms/mole).
This type of problem is common in chemistry for practicing mole and Avogadro's number calculations.
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