<h3>Answer:</h3>
64 g O₂
<h3>General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] CH₄ + 2O₂ → CO₂ + 2H₂O
[Given] 36 g H₂O
[Solve] x g O₂
<u>Step 2: Identify Conversions</u>
[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
<u>Step 3: Stoichiometry</u>
- Set up conversion:

- Divide/Multiply [Cancel Units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
63.929 g O₂ ≈ 64 g O₂
Answer: 0.4 moles
Explanation:
Given that:
Volume of gas V = 11L
(since 1 liter = 1dm3
11L = 11dm3)
Temperature T = 25°C
Convert Celsius to Kelvin
(25°C + 273 = 298K)
Pressure P = 0.868 atm
Number of moles N = ?
Note that Molar gas constant R is a constant with a value of 0.00821 atm dm3 K-1 mol-1
Then, apply ideal gas equation
pV = nRT
0.868atm x 11dm3 = n x (0.00821 atm dm3 K-1 mol-1 x 298K)
9.548 atm dm3 = n x 24.47atm dm3mol-1
n = (9.548 atm dm3 / 24.47atm dm3 mol-1)
n = 0.4 moles
Thus, there are 0.4 moles of the gas.
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