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How many grams of CO2 are produced from 6.7 L of O2 gas at STP?

1 Answer

3 votes

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

Step-by-step 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:
    \displaystyle 6.7 \ L \ O_2((1 \ mol \ O_2)/(22.4 \ L \ O_2))((44.01 \ g \ O_2)/(1 \ mol \ O_2))
  2. [DA] Divide/Multiply [Cancel out units]:
    \displaystyle 13.1637 \ g \ O_2

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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