Question

CH4 + 202 → CO2 + 2H2O How many grams of O2 needed to produce 36 grams of H2O?

Answer 1

Answer:

64 g 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]   36 g H₂O

[Solve]   x g O₂

Step 2: Identify Conversions

[RxN] 2 mol O₂ → 2 mol H₂O

[PT] Molar Mass of O - 16.00 g/mol

[PT] Molar Mas of H - 1.01 g/mol

Molar Mass of O₂ - 2(16.00) = 32.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

Step 3: Stoichiometry

1. Set up conversion:                     $\displaystyle 36 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O})(\frac{2 \ mol \ O_2}{2 \ mol \ H_2O})(\frac{32.00 \ g \ O_2}{1 \ mol \ O_2})$
2. Divide/Multiply [Cancel Units]:                                                                       $\displaystyle 63.929 \ g \ O_2$

Step 4: Check

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

63.929 g O₂ ≈ 64 g O₂

Answer 2

Mass of O₂ needed = 64 g

Further explanation

Given

Reaction

CH₄ + 2O₂ → CO₂ + 2H₂O

36 grams of H₂O

Required

Mass of O₂ needed

Solution

mol H₂O :

= mass : MW H₂O

= 36 : 18 g/mol

= 2

From the equation, mol ratio of O₂ : H₂O = 2 : 2, so mol O₂=2

Mass O₂ :

= mol x MW O₂

= 2 x 32 g/mol

= 64 g