Question

Stoichiometry is hawd

Using the following reaction:

C12H22O11 + 12 O2 → 12CO2 + 11H2O + energy

How many grams of carbon dioxide, CO2, are produced from metabolizing 448g sucrose?

Answer 1

691 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
• Moles

Organic

• Organic Compounds CₓHₙ

Stoichiometry

• Using Dimensional Analysis
• Analyzing Reactions RxN

Step-by-step explanation:

Step 1: Define

[RxN - Balanced] C₁₂H₂₂O₁₁ + 12O₂ → 12CO₂ + 11H₂O + Energy

[Given] 448 g C₁₂H₂₂O₁₁ (sucrose)

[Solve] grams CO₂ (carbon dioxide)

Step 2: Identify Conversions

[RxN] 1 mol C₁₂H₂₂O₁₁ → 12 mol CO₂

[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 C₁₂H₂₂O₁₁ - 12(12.01) + 22(1.01) + 11(16.00) = 342.34 g/mol

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

Step 3: Stoich

1. [DA] Set up conversion:
   `\displaystyle 448 \ g \ C_(12)H_(22)O_(11)((1 \ mol \ C_(12)H_(22)O_(11))/(342.34 \ g \ C_(12)H_(22)O_(11)))((12 \ mol \ CO_2)/(1 \ mol \ C_(12)H_(22)O_(11)))((44.01 \ g \ CO_2)/(1 \ mol \ CO_2))`
2. [DA] Multiply/Divide [Cancel out units]:
   `\displaystyle 691.119 \ g \ CO_2`

Step 4: Check

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

691.119 g CO₂ ≈ 691 g CO₂