Question

CH4 + O2 → CO2 + H2O
How many molecules of water can be formed by the combustion of 130 g of methane?

Answer 1

Answer:

9.8 × 10²⁴ molecules H₂O

General Formulas and Concepts:

Atomic Structure

• Reading a Periodic Table
• Moles
• Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Organic

• Naming carbons

Stoichiometry

• Analyzing reaction rxn
• Using Dimensional Analysis

Explanation:

Step 1: Define

[RxN - Unbalanced] CH₄ + O₂ → CO₂ + H₂O

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

[Given] 130 g CH₄

Step 2: Identify Conversions

Avogadro's Number

[RxN] 1 mol CH₄ → 2 mol H₂O

[PT] Molar Mass of C: 12.01 g/mol

[PT] 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. [DA] Set up conversion:                                                                                   $\displaystyle 130 \ g \ CH_4(\frac{1 \ mol \ CH_4}{16.05 \ g \ CH_4})(\frac{2 \ mol \ H_2O}{1 \ mol \ CH_4})(\frac{6.022 \cdot 10^{23} \ molecules \ H_2O}{1 \ mol \ H_2O})$
2. [DA] Divide/Multiply [Cancel out units]:                                                           $\displaystyle 9.75526 \cdot 10^{24} \ molecules \ H_2O$

Step 4: Check

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

9.75526 × 10²⁴ molecules H₂O ≈ 9.8 × 10²⁴ molecules H₂O