<h3>Answer:</h3>
60 g O₂
<h3>General Formulas and Concepts: </h3><u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
<u>Stoichiometry</u>
- Using Dimensional Analysis
<u>Step 1: Define</u>
[RxN - Balanced] CH₄ + 2O₂ → CO₂ + 2H₂O
[Given] 2 mol H₂O
[Solve] x g O₂
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol H₂O → 2 mol O₂
[PT] Molar Mass of O - 16.00 g/mol
Molar Mass of O₂ - 2(16.00) = 32.00 g/mol
<u>Step 3: Stoichiometry</u>
- Set up conversion:
- Divide/Multiply:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 1 sig fig.</em>
64.00 g O₂ ≈ 60 g O₂