Answer:
290 g Al₂O₃
General Formulas and Concepts:
Atomic Structure
- Reading a Periodic Table
- Moles
Stoichiometry
- Using Dimensional Analysis
Explanation:
Step 1: Define
[Given] 2.8 mol Al₂O₃
[Solve] g Al₂O₃
Step 2: Identify Conversions
[PT] Molar Mass of Al: 26.98 g/mol
[PT] Molar Mass of O: 16.00 g/mol
Molar Mass of Al₂O₃: 2(26.98) + 3(16.00) = 101.96 g/mol
Step 3: Convert
- [DA] Set up: [tex]\displaystyle 2.8 \ mol \ Al_2O_3(\frac{101.96 \ g \ Al_2O_3}{1 \ mol \ Al_2O_3})[/tex]
- [DA] Multiply [Cancel out units]: [tex]\displaystyle 285.488 \ g \ Al_2O_3[/tex]
Step 4: Check
Follow sig fig rules and round. We are given 2 sig figs.
285.488 g Al₂O₃ ≈ 290 g Al₂O₃
Topic: AP Chemistry
Unit: Atomic Structure
