Ch.3 - Chemical ReactionsWorksheetSee 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: For the reaction 2C2H5OH + 2Na2Cr2O7 + 16HCI → 3CH3COOH + 4CrCl3 + 4NaCl + 11H2O the yield (called a theoretical yield) of CrCl3 we can get from the reaction of 14.0 mol of C2H5OH, 10.0 mol of Na2Cr2O

Problem

For the reaction 

2C2H5OH + 2Na2Cr2O7 + 16HCI → 3CH3COOH + 4CrCl3 + 4NaCl + 11H2

the yield (called a theoretical yield) of CrCl3 we can get from the reaction of 14.0 mol of C2H5OH, 10.0 mol of Na2Cr2O7 and 50.0 mol of HCI would be 

(a) 20.0 mol 

(b) 28.0 mol 

(c) 10.0 mol 

(d) 12.5 mol