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

64 g O₂

Math

Pre-Algebra

Order of Operations: BPEMDAS

Chemistry

Atomic Structure

Stoichiometry

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

Step 4: Check

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

63.929 g O₂ ≈ 64 g O₂

Mass of O₂ needed = 64 g

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

a galvanic cell( aka voltaic cell) and electrolytic cell

explanation:

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