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: In the reaction of 1.00 mole of O2 with 2.00 mole of H2, 2H2  + O2→ 2H2O If 30.0g of H2O is formed, what is the percent yield?a)  100%                  b)  88.9%              c)  50.0%              d)  167%                e)  83.3%

Problem

In the reaction of 1.00 mole of O2 with 2.00 mole of H2, 2H2  + O2→ 2H2O

 If 30.0g of H2O is formed, what is the percent yield?

a)  100%                  

b)  88.9%              

c)  50.0%              

d)  167%                

e)  83.3%

Solution

We’re being asked to calculate the percent yield of H2O. The balanced chemical equation is:


2 H2 + O2 → 2 H2O


Recall that percent yield is given by:


% Yield = actual yieldtheoretical yield × 100


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