Respuesta :
Answer:
60 g O₂
General Formulas and Concepts:
Math
Pre-Algebra
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- 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] 2 mol H₂O
[Solve] x g O₂
Step 2: Identify Conversions
[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
Step 3: Stoichiometry
- Set up conversion: [tex]\displaystyle 2 \ mol \ H_2O(\frac{2 \ mol \ O_2}{2 \ mol \ H_2O})(\frac{32.00 \ g \ O_2}{1 \ mol \ O_2})[/tex]
- Divide/Multiply: [tex]\displaystyle 64.00 \ g \ O_2[/tex]
Step 4: Check
Follow sig fig rules and round. We are given 1 sig fig.
64.00 g O₂ ≈ 60 g O₂
Mass of O₂= 64 g
Further explanation
Given
Reaction
CH₄ + 20₂ → CO₂ + 2H₂O
2 moles of H₂O
Required
Mass of O₂
Solution
In a chemical equation, the reaction coefficient shows the mole ratio of the compounds involved in the reaction, reactants or products
From the equation, mol ratio of H₂O : O₂ = 1 : 1, so moles O₂ :
= 1/1 x moles H₂O
= 1/1 x 2
= 2 moles
Mass of O₂ :
= mol x MW
= 2 x 32 g/mol
= 64 g