Question

How many grams of Fe will be produced if 705.0g of H20 are produced?

Answer 1

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:                              $\displaystyle 705.0 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O})(\frac{1 \ mol \ Fe_3O_4}{4 \ mol \ H_2O})(\frac{231.55 \ g \ Fe_3O_4}{1 \ mol \ Fe_3O_4})$
2. Multiply/Divide/Cancel units:                                                                           $\displaystyle 2264.74 \ g \ Fe_3O_4$

Step 4: Check

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

2264.74 g Fe₃O₄ ≈ 2265 g Fe₃O₄