Question

How many g of Al are required to produce 2.8 mol of Al2O3

Answer 1

290 g is the answer yeaaaa

Answer 2

290 g Al₂O₃

General Formulas and Concepts:

Atomic Structure

• Reading a Periodic Table
• Moles

Stoichiometry

• Using Dimensional Analysis

Step-by-step 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

1. [DA] Set up:
   `\displaystyle 2.8 \ mol \ Al_2O_3((101.96 \ g \ Al_2O_3)/(1 \ mol \ Al_2O_3))`
2. [DA] Multiply [Cancel out units]:
   `\displaystyle 285.488 \ g \ Al_2O_3`

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