Answer:
2265 g Fe₃O₄
General Formulas and Concepts:
Math
Pre-Algebra
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Chemistry
Atomic Structure
Stoichiometry
- Using Dimensional Analysis
Step-by-step 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
- Set up stoich:
![\displaystyle 705.0 \ g \ H_2O((1 \ mol \ H_2O)/(18.02 \ g \ H_2O))((1 \ mol \ Fe_3O_4)/(4 \ mol \ H_2O))((231.55 \ g \ Fe_3O_4)/(1 \ mol \ Fe_3O_4))](https://img.qammunity.org/2022/formulas/chemistry/college/5ece07lztyd1zha1qbwhp0p7gf0hv25777.png)
- Multiply/Divide/Cancel units:
![\displaystyle 2264.74 \ g \ Fe_3O_4](https://img.qammunity.org/2022/formulas/chemistry/college/yvoyq16k1amgmyvth6wcjb8liemogg63hm.png)
Step 4: Check
Follow sig fig rules and round. We are given 4 sig figs.
2264.74 g Fe₃O₄ ≈ 2265 g Fe₃O₄