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: What is the percent yield for CH3Br if 12.23 g is produced when 5.00 g of CH4O reacts with excess HBr? CH4O + HBr → CH3Br + H2OA. 16% B. 67% C. 84% D. 82.6% E. 12%

Problem

What is the percent yield for CH3Br if 12.23 g is produced when 5.00 g of CH4O reacts with excess HBr? CH4O + HBr → CH3Br + H2O

A. 16% 

B. 67% 

C. 84% 

D. 82.6% 

E. 12%

Solution

We’re being asked to calculate the percent yield of CH3Br in the reaction of 5.00 g HBr with excess CH4O. The balanced chemical equation is:

 CH4O + HBr → CH3Br + H2O


Recall that percent yield is given by:


%yield = actual yieldtheoretical yieldx100



We know the reaction produced 12.23 g CH3Br; this is the actual yield of the reaction. To calculate the theoretical yield, we need to do the following:


Mass of CH4(molar mass of CH4O)  Moles of CH4(mole-to-mole comparison)  Moles of CH3Br (molar mass of CH3Br)  Mass of CH3Br


Since HBr is in excess, we can simply ignore it in our calculations.


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