Respuesta :

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Answer:

4.0 mol O₂

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Chemistry

Atomic Structure

  • Reading a Periodic Table

Stoichiometry

  • Using Dimensional Analysis

Explanation:

Step 1: Define

[RxN - Balanced]   CH₄ + 2O₂ → CO₂ + 2H₂O

[Given]   32 g CH₄

Step 2: Identify Conversions

[RxN] 1 mol CH₄ → 2 mol O₂

Molar Mass of C - 12.01 g/mol

Molar Mass of H - 1.01 g/mol

Molar Mass of CH₄ - 12.01 + 4(1.01) = 16.05 g/mol

Step 3: Stoichiometry

  1. Set up conversion:                    [tex]\displaystyle 32 \ g \ CH_4(\frac{1 \ mol \ CH_4}{16.05 \ g \ CH_4})(\frac{2 \ mol \ O_2}{1 \ mol \ CH_4})[/tex]
  2. Divide/Multiply:                          [tex]\displaystyle 3.98754 \ mol \ O_2[/tex]

Step 4: Check

Follow sig fig rules and round. We are given 2 sig figs.

3.98754 mol O₂ ≈ 4.0 mol O₂

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