Answer:
5.595 g Fe
General Formulas and Concepts:
Chemistry - Stoichiometry
- Reading a Periodic Table
- Using Dimensional Analysis
Explanation:
Step 1: Define
16.25 g FeCl₃ (Iron III Chloride)
RxN: 2Fe + 3Cl₂ → 2FeCl₃
Step 2: Identify Conversions
Molar Mass of Fe - 55.85 g/mol
Molar Mass of Cl - 35.45 g/mol
Molar Mass of FeCl₃ - 55.85 + 3(35.45) = 162.2 g/mol
Step 3: Stoichiometry
[tex]16.25 \ g \ FeCl_3(\frac{1 \ mol \ FeCl_3}{162.2 \ g \ FeCl_3} )(\frac{2 \ mol \ Fe}{2 \ mol FeCl_3} )(\frac{55.85 \ g \ Fe}{1 \ mol \ Fe} )[/tex] = 5.59533 g Fe
Step 4: Check
We are given 4 sig figs. Follow sig fig rules and round.
5.59533 g Fe ≈ 5.595 g Fe
