Ch.3 - Chemical ReactionsSee all chapters
All Chapters
Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds

Solution: When 4.50 g of Fe2O3 is reduced with excess H2 in a furnace, 2.60 g of metallic iron is recovered. What is the percent yield? This is the equation representing the reaction. Fe2O3 has a molar mass of 159.7 g x mol-1 Fe2O3 (s) + 3 H2 (g) → 2 Fe (s) + 3 H2O (g) a) 82.6% b) 70.0% c) 57.8% d) 31.5%

Problem

When 4.50 g of Fe2O3 is reduced with excess H2 in a furnace, 2.60 g of metallic iron is recovered. What is the percent yield? This is the equation representing the reaction. Fe2O3 has a molar mass of 159.7 g x mol-1

Fe2O(s) + 3 H(g) → 2 Fe (s) + 3 H2O (g)

a) 82.6%

b) 70.0%

c) 57.8%

d) 31.5%