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

Sagot :

Space

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

  1. [DA] Set up:                                                                                                       [tex]\displaystyle 2.8 \ mol \ Al_2O_3(\frac{101.96 \ g \ Al_2O_3}{1 \ mol \ Al_2O_3})[/tex]
  2. [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

jeramymp
290 g is the answer yeaaaa
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