Respuesta :

Space

Answer:

120 g CO₂

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] C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

[Given] 150 g O₂

Step 2: Identify Conversions

[RxN] 5 mol O₂ → 3 mol CO₂

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

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

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

Molar Mass of CO₂ - 12.01 + 2(16.00) = 44.01 g/mol

Step 3: Stoich

  1. [DA] Set up:                                                                                                     [tex]\displaystyle 150 \ g \ O_2(\frac{1 \ mol \ O_2}{32.00 \ g \ O_2})(\frac{3 \ mol \ CO_2}{5 \ mol \ O_2})(\frac{44.01 \ g \ CO_2}{1 \ mol \ CO_2})[/tex]
  2. [DA] Multiply/Divide [Cancel out units]:                                                         [tex]\displaystyle 123.778 \ g \ CO_2[/tex]

Step 4: Check

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

123.778 g CO₂ ≈ 120 g CO₂

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