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erastovalidia [21]
3 years ago
11

CH4 + 202 → CO2 + 2H2O How many grams of O2 needed to react with 2 moles of CH4?

Chemistry
2 answers:
inn [45]3 years ago
4 0

Mass of O₂= 128 g

<h3>Further explanation</h3>

Given

Reaction

CH₄ + 20₂ → CO₂ + 2H₂O

2 moles of CH₄

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 CH₄ : O₂ = 1 : 2, so moles O₂ :

= 2/1 x moles CH₄

= 2/1 x 2

= 4 moles

Mass of O₂ :

= mol x MW

= 4 x 32 g/mol

= 128 g

PtichkaEL [24]3 years ago
3 0
<h3>Answer:</h3>

100 g O₂

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

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

[Given]   2 mol CH₄

[Solve]   x g O₂

<u>Step 2: Identify Conversions</u>

[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

<u>Step 3: Stoichiometry</u>

  1. Set up conversion:                     \displaystyle 2 \ mol \ CH_4(\frac{2 \ mol \ O_2}{1 \ mol \ CH_4})(\frac{32.00 \ g \ O_2}{1 \ mol \ O_2})
  2. Multiply/Divide:                           \displaystyle 128 \ g \ O_2

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

128 g O₂ ≈ 100 g O₂

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