Ch.6 - Thermochemistry WorksheetSee 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: The reaction for the combustion of propane is shown below. C3H8(g) + 5 O2 (g) → 3 CO2 (g)  + 4 H2O (l)    ΔH = - 526.3 kcal/molHow much heat is  released when 71.7 g of propane is burned? 

Solution: The reaction for the combustion of propane is shown below. C3H8(g) + 5 O2 (g) → 3 CO2 (g)  + 4 H2O (l)    ΔH = - 526.3 kcal/molHow much heat is  released when 71.7 g of propane is burned? 

Problem

The reaction for the combustion of propane is shown below. 

C3H8(g) + 5 O2 (g) → 3 CO2 (g)  + 4 H2O (l)    ΔH = - 526.3 kcal/mol

How much heat is  released when 71.7 g of propane is burned? 

Solution

First, we have to take note that we were given the standard enthalpy of combustion for  C3H8(g)  and this is for 1 mole of C3H8(g). 

We are given the mass so we need to convert that first to moles in order to calculate. 

View the complete written solution...