Chapter 3: Chemical Reactions and Reaction Stoichiometry

Stoichiometry

Stoichiometry is defined as the study of mass relationships in chemistry. This field is based on the Law of Conservation of Mass, which was articulated by Antoine Lavoisier in 1789. This fundamental axiom states that in all operations of art and nature, nothing is created, meaning an equal amount of matter exists both before and after an experiment. The entire art of performing chemical experiments depends upon this principle.

Chemical Equations

Chemical equations are concise representations of chemical reactions. These equations consist of reactants, which are the substances that undergo the reaction and are found on the left side of the equation, and products, which are the substances produced by the reaction and are located on the right side. An example of a chemical reaction equation is: 2H2+O2→2H2O2H2+O2→2H2​O

Key Elements in Chemical Equations

Chemical equations include state symbols, which indicate the physical state of reactants and products, typically enclosed in parentheses. These symbols include (g) for gas, (l) for liquid, (s) for solid, and (aq) for an aqueous solution. Balancing chemical equations is crucial as it requires coefficients to ensure the same number of each type of atom on both sides of the equation, thereby reflecting the conservation of mass.

Why Use Coefficients Instead of Changing Subscripts?

It is important to use coefficients for balancing chemical equations instead of changing subscripts because altering subscripts changes the actual compounds formed. For instance, 2H2+O2→2H2O2H2+O2→2H2O produces water, whereas if subscripts were changed, H2+O2→H2O2H2+O2→H2O2​ would produce hydrogen peroxide, a different substance entirely.

Balancing Examples

Here are some exercises to practice balancing chemical equations:

• Exercise 1: Balance the reaction between methane and bromine. The unbalanced equation is: CH4(g)+Br2(l)→CBr4(s)+HBr(g)CH4(g)+Br2(l)→CBr4​(s)+HBr(g)

• Exercise 2: Provide missing coefficients for the following reactions:

  • (a) Fe(s)+O2(g)→Fe2O3(s)Fe(s)+O2(g)→Fe2O3​(s)

  • (b) Al(s)+HCl(aq)→AlCl3(aq)+H2(g)Al(s)+HCl(aq)→AlCl3(aq)+H2(g)

  • (c) CaCO3(s)+HCl(aq)→CaCl2(aq)+CO2(g)+H2O(l)CaCO3(s)+HCl(aq)→CaCl2(aq)+CO2(g)+H2O(l)

Types of Reactions

There are three main types of chemical reactions:

1. Combination Reactions: In this type, two or more substances combine to form a single product. Examples include 2Mg(s)+O2(g)→2MgO(s)2Mg(s)+O2(g)→2MgO(s), N2(g)+3H2(g)→2NH3(g)N2(g)+3H2(g)→2NH3(g), and C3H6(g)+Br2(l)→C3H6Br2(l)C3H6(g)+Br2(l)→C3H6Br2(l).

2. Decomposition Reactions: These reactions involve one substance breaking down into two or more simpler substances. Examples are CaCO3(s)→CaO(s)+CO2(g)CaCO3(s)→CaO(s)+CO2(g), 2KClO3(s)→2KCl(s)+O2(g)2KClO3(s)→2KCl(s)+O2(g), and 2NaN3(s)→2Na(s)+3N2(g)2NaN3(s)→2Na(s)+3N2(g).

3. Combustion Reactions: These are rapid reactions that typically produce heat and light and always involve oxygen. Common examples include CH4(g)+2O2(g)→CO2(g)+2H2O(g)CH4(g)+2O2(g)→CO2(g)+2H2O(g) and C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)C3H8(g)+5O2(g)→3CO2(g)+4H2​O(g).

General rules for combustion state that organic compounds containing carbon, hydrogen, and oxygen combust with oxygen to produce carbon dioxide (CO2CO2) and water (H2OH2O).

Balanced Equation for Combustion of Methanol

Consider the combustion of methanol. Starting with the unbalanced equation: 2CH3OH(l)+O2(g)→CO2(g)+H2O(g)2CH3OH(l)+O2(g)→CO2(g)+H2O(g). After arranging coefficients to balance the equation, it appears as: 2CH3OH(l)+3O2(g)→2CO2(g)+4H2O(g)2CH3OH(l)+3O2(g)→2CO2(g)+4H2O(g).

Formula Weight and Molecular Weight

Formula Weight (FW) is the sum of the atomic weights of all elements in a compound. For example, for CaCl2CaCl2, the formula weight is calculated as 1(40.08 amu)+2(35.453 amu)=110.99 amu1(40.08 amu)+2(35.453 amu)=110.99 amu. Molecular Weight (MW) is specifically the sum of atomic weights of a molecule’s atoms. For C2H6C2H6​, the molecular weight is 2(12.011 amu)+6(1.00794 amu)=30.070 amu+6.04764 amu=36.12 amu2(12.011 amu)+6(1.00794 amu)=30.070 amu+6.04764 amu=36.12 amu.

Ionic Compounds and Formulas

Ionic compounds are structured in a three-dimensional order, which means that their composition is better represented by empirical formulas rather than molecular formulas.

Percent Composition

To determine the percentage by mass of an element in a compound, the Percent Composition equation is used: Percent Composition=(number of atoms)×(atomic weight)FW of the compound×100 Percent Composition=FW of the compound(number of atoms)×(atomic weight)​×100. For instance, one can calculate the percentage of carbon in a given compound using this formula.

Example Calculation for Sucrose (C12H22O11)

First, determine the formula weight of sucrose, C12H22O11C12H22O11, which totals 342.0 amu342.0 amu. As a practice example, find the formula weight of Ca(NO3)2Ca(NO3)2​ given its atomic constituents.

Avogadro’s Number

Avogadro’s Number is defined as 6.02×10236.02×1023 atoms or molecules, which constitutes one mole. This number is significant in laboratory work because moles act as a bridge between the molecular scale and practical measurements in active laboratory settings.

Molar Mass

Molar mass is defined as the mass of 1 mole of a substance, typically expressed in grams per mole (g/mol). It is numerically equal to the formula weight when expressed in atomic mass units (amu).

Stoichiometric Calculations

The coefficients within balanced chemical equations indicate the relationships of moles in reactions, which can then be related to the mass of reactants and products.

Limiting Reactants

The limiting reactant is defined as the reactant that is entirely consumed first, thereby limiting the overall extent of the reaction. When determining the yield of a reaction, it is always important to consider the limiting reactant.

Practice Problem

For a given reaction, N2(g)+3H2(g)→2NH3(g)N2(g)+3H2(g)→2NH3(g), determine how many moles of ammonia (NH3NH3) can be produced from provided quantities of nitrogen (N2N2) and hydrogen (H2H2).

Theoretical Yield and Percent Yield

Theoretical Yield refers to the maximum amount of product that can be feasibly produced according to stoichiometric calculations. Percent Yield is then calculated using the equation: Percent Yield=actual yieldtheoretical yield×100Percent Yield=theoretical yieldactual yield​×100. An example calculation involves determining the theoretical yield and percent yield from specific limiting reactant conditions in an experiment with known outputs.

Practice Examples

Practice examples often involve balancing equations and performing various stoichiometric calculations across multiple reagents and conditions. When working with these examples, it's crucial to identify the limiting reactant first, as it will dictate the maximum theoretical yield of the desired product. After calculating the theoretical yield based on the limiting reactant, students can then compare it to the actual yield obtained from their experiments to compute the percent yield, thus assessing the efficiency of the reaction. Furthermore, this process not only reinforces their understanding of stoichiometry but also enhances their laboratory skills by prompting them to critically analyze their results and consider factors affecting yield, such as purity of reagents and measurement accuracy.

Chemical Equations

• 2H₂(g)+O₂(g)→2H_2​O(l)

• H₂+O₂→H₂O₂

Balancing Examples

• Exercise 1: CH₄(g)+Br₂(l)→CBr₄​(s)+HBr(g)

• Exercise 2:

  • (a) Fe(s)+O₂(g)→Fe₂O₃​(s)

  • (b) Al(s)+HCl(aq)→AlCl₃(aq)+H₂(g)

  • (c) CaCO₃(s)+HCl(aq)→CaCl₂(aq)+CO₂(g)+H2O(l)

Types of Reactions

1. Combination Reactions:

   • $2Mg(s)+O_2(g)\rightarrow2MgO(s)$

   • N₂(g)+3H₂(g)→2NH₃​(g)

   • C₃H₆(g)+Br₂(l)→C₃H₆Br₂(l)

2. Decomposition Reactions:

   • CaCO₃(s)→CaO(s)+CO₂(g)

   • 2KClO₃(s)→2KCl(s)+O₂(g)

   • 2NaN₃(s)→2Na(s)+3N₂(g)

3. Combustion Reactions:

   • CH4(g)+2O2(g)→CO2(g)+2H2O(g)

   • C₃H₈(g)+5O₂(g)→3CO₂(g)+4H₂​O(g)

General rules for combustion

• $CO_2$

• $H_2O$

Balanced Equation for Combustion of Methanol

• Unbalanced: CH₃OH(l)+O₂(g)→CO₂(g)+H2O(g)

• Balanced: 2CH₃OH(l)+3O₂(g)→2CO₂(g)+4H2O(g)

Formula Weight and Molecular Weight

• Formula Weight example: $CaCl_2$

  • Calculation: $1(40.08\ amu)+2(35.453\ amu)=110.99\ amu$

• Molecular Weight example: C₂H₆

  • Calculation: $2(12.011\ amu)+6(1.00794\ amu)=30.070\ amu+6.04764\ amu=36.12\ amu$

Percent Composition

• Formula: $\text{Percent Composition}=\frac{\text{(number of atoms)}\times(\text{atomic weight})}{\text{FW of the compound}}\times100$

Example Calculation for Sucrose (C₁₂H₂₂O₁₁​)

• Formula: C₁₂H₂₂O_11​
  

• Formula weight: $342.0\ amu$

• Practice example: Ca(NO3)₂

Avogadro’s Number

• $6.02\times10^{23}$

Limiting Reactants Practice Problem

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• Reaction: N₂(g)+3H₂(g)→2NH_3​(g)

• Compound examples: NH₃$,$ N₂$,$H₂

Theoretical Yield and Percent Yield

• Formula: $\text{Percent Yield}=\frac{\text{actual yield}}{\text{theoretical yield}}\times100$