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: Give the percent yield when 28.16 g of CO2 (molar mass = 44.01 g/mol) are formed from the reaction of 4.000 moles of C8H18 with 4.000 moles of O2.2 C8H18 + 25 O2 → 16 CO2 + 18 H2OA) 20.00% B) 25.00% C

Problem

Give the percent yield when 28.16 g of CO2 (molar mass = 44.01 g/mol) are formed from the reaction of 4.000 moles of C8H18 with 4.000 moles of O2.

2 C8H18 + 25 O→ 16 CO2 + 18 H2O

A) 20.00% 

B) 25.00% 

C) 50.00% 

D) 12.50%

Solution

Use the concept of limiting reactant to determine the theoretical yield of CO2 when 4 moles of C8H8 and 4 moles of O2 reacts.

The given reaction is already balanced, all we need is the molar mass of CO2:

CO2        1 C × 12.01 g/mol C = 12.01 g/mol
               2 O × 16 g/mol O = 32 g/mol
                       Sum 
44.01 g/mol

View the complete written solution...