Answer:
100 g O₂
General Formulas and Concepts:
Math
Pre-Algebra
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Chemistry
Atomic Structure
Stoichiometry
- Using Dimensional Analysis
Step-by-step explanation:
Step 1: Define
[RxN - Balanced] CH₄ + 2O₂ → CO₂ + 2H₂O
[Given] 2 mol CH₄
[Solve] x g O₂
Step 2: Identify Conversions
[RxN] 1 mol CH₄ → 2 mol O₂
[PT] Molar Mass of O - 16.00 g/mol
Molar Mass of O₂ - 2(16.00) = 32.00 g/mol
Step 3: Stoichiometry
- Set up conversion:
![\displaystyle 2 \ mol \ CH_4((2 \ mol \ O_2)/(1 \ mol \ CH_4))((32.00 \ g \ O_2)/(1 \ mol \ O_2))](https://img.qammunity.org/2022/formulas/chemistry/high-school/8ys3iyws32c55oh0hzrbx9mtxcymy9t94y.png)
- Multiply/Divide:
![\displaystyle 128 \ g \ O_2](https://img.qammunity.org/2022/formulas/chemistry/college/h8v7vxzzyz8s2n7svl432f2y26i25opw3z.png)
Step 4: Check
Follow sig fig rules and round. We are given 1 sig fig.
128 g O₂ ≈ 100 g O₂