Respuesta :

Space

Answer:

13 g CO₂

General Formulas and Concepts:

Chemistry

Atomic Structure

  • Reading a Periodic Table
  • Moles
  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

Stoichiometry

  • Using Dimensional Analysis

Math

Pre-Algebra

Order of Operations: BPEMDAS

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

Explanation:

Step 1: Define

Identify variables

[Given] 6.7 L O₂

[Solve] g O₂

Step 2: Identify Conversions

[STP] 22.4 L = 1 mol

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

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

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

Step 3: Convert

  1. [DA] Set up:                                                                                                       [tex]\displaystyle 6.7 \ L \ O_2(\frac{1 \ mol \ O_2}{22.4 \ L \ O_2})(\frac{44.01 \ g \ O_2}{1 \ mol \ O_2})[/tex]
  2. [DA] Divide/Multiply [Cancel out units]:                                                           [tex]\displaystyle 13.1637 \ g \ O_2[/tex]

Step 4: Check

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

13.1637 g CO₂ ≈ 13 g CO₂

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