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Answer:

2265 g Fe₃O₄

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] Fe₃O₄ + 4H₂ → 3Fe + 4H₂O

[Given] 705.0 g H₂O

Step 2: Identify Conversions

[RxN] 4 mol H₂O → 1 mol Fe₃O₄

Molar Mass of H - 1.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Fe - 55.85 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

Molar Mass of Fe₃O₄ - 3(55.85) + 4(16.00) = 231.55 g/mol

Step 3: Convert

  1. Set up stoich:                              [tex]\displaystyle 705.0 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O})(\frac{1 \ mol \ Fe_3O_4}{4 \ mol \ H_2O})(\frac{231.55 \ g \ Fe_3O_4}{1 \ mol \ Fe_3O_4})[/tex]
  2. Multiply/Divide/Cancel units:                                                                           [tex]\displaystyle 2264.74 \ g \ Fe_3O_4[/tex]

Step 4: Check

Follow sig fig rules and round. We are given 4 sig figs.

2264.74 g Fe₃O₄ ≈ 2265 g Fe₃O₄

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