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Answer:

5.55 mol C₂H₅OH

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 Tables
  • Moles

Stoichiometry

  • Using Dimensional Analysis
  • Analyzing Reactions RxN

Explanation:

Step 1: Define

[RxN - Balanced] C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

[Given] 500. g C₆H₁₂O₆ (Glucose)

[Solve] moles C₂H₅OH (Ethanol)

Step 2: Identify Conversions

[RxN] 1 mol C₆H₁₂O₆ → 2 mol C₂H₅OH

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

[PT] Molar Mass of H - 1.01 g/mol

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

Molar Mass of C₆H₁₂O₆ - 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol

Step 3: Stoichiometry

  1. [DA] Set up conversion:                                                                                 [tex]\displaystyle 500 \ g \ C_6H_{12}O_6(\frac{1 \ mol \ C_6H_{12}O_6}{180.18 \ g \ C_6H_{12}O_6})(\frac{2 \ mol \ C_2H_5OH}{1 \ mol \ C_6H_{12}O_6})[/tex]
  2. [DA} Multiply/Divide [Cancel out units]:                                                         [tex]\displaystyle 5.55001 \ mol \ C_2H_5OH[/tex]

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

5.55001 mol C₂H₅OH ≈ 5.55 mol C₂H₅OH

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