General Chemistry: Reaction Types, Balancing Equations, Stoichiometry, and Limiting Reactants

Reaction Types

Ionic and Redox Reactions

Chemical reactions are classified by the types of changes that occur to atoms and molecules. Understanding these types is essential for predicting products and balancing equations.

• Ionic Reactions: Involve the exchange of ions between reactants. Common types include double displacement and acid/base reactions.

• Redox Reactions: Involve the transfer of electrons. Balancing often requires separating into half-reactions.

• Combustion: A reaction where a substance combines with oxygen to produce carbon dioxide and water. $\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$

• Single Replacement: An element replaces another in a compound.

• Decomposition: A compound breaks down into simpler substances.

Balancing Chemical Equations

By Inspection

Balancing chemical equations ensures the conservation of mass and atoms. The process is iterative and requires adjusting coefficients so that the number of atoms of each element is equal on both sides.

• Rules:

  • Conservation of mass and atoms

  • Change coefficients, not subscripts

  • Include phases (solid, liquid, gas, aqueous)

• Order of Operation:

  1. Write reactants on the left, products on the right.

  2. Ensure correct subscripts and charges for all species.

  3. Balance elements with odd numbers last.

  4. Balance free elements last.

  5. Iteratively adjust coefficients until balanced.

  6. Reduce to smallest whole number ratios.

Example: $\text{CH}_3\text{CHO} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$ Balance by inspection. *Additional info: Polyatomic ions can be balanced as units to simplify the process.*

Stoichiometry and Mole Calculations

Using Moles to Relate Quantities

Stoichiometry allows chemists to predict the amounts of reactants and products in a chemical reaction using the concept of the mole.

• Mole Map:

  grams	Mol A	mol ratio	Mol B	grams
  grams	Mol A	mol ratio	Mol B	grams

  $M = \text{molar mass}$ $A = \text{Avogadro's number}$

• Example: $4\text{NH}_3 (g) + 5\text{O}_2 (g) \rightarrow 4\text{NO} (g) + 6\text{H}_2\text{O} (l)$ If you have 10 mol of $\text{O}_2$, how many mol $\text{NH}_3$ are needed? $\frac{4\text{mol NH}_3}{5\text{mol O}_2} \times 10\text{mol O}_2 = 8\text{mol NH}_3$ How many mol $\text{H}_2\text{O}$ are generated? $\frac{6\text{mol H}_2\text{O}}{5\text{mol O}_2} \times 10\text{mol O}_2 = 12\text{mol H}_2\text{O}$

Limiting Reactants in Stoichiometry Problems

Identifying the Limiting Reactant

The limiting reactant is the substance that is completely consumed first, limiting the amount of product formed. This concept is crucial for predicting yields in chemical reactions.

• Compare the mole ratios of reactants to the coefficients in the balanced equation.

• The reactant that produces the least amount of product is the limiting reactant.

• Example: If a bicycle assembly requires 1 frame and 2 wheels, and you have 4 frames and 8 wheels, you can make 4 bicycles. If you have 4 frames and only 2 wheels, you can make only 1 bicycle.

Precipitation Reactions and Tables

Tracking Ion Changes and Product Formation

Precipitation reactions involve the formation of an insoluble product (precipitate) from soluble reactants. Tracking the moles of ions before and after the reaction helps determine the amount of product formed.

• Use molecular views and tables to summarize initial moles, changes, and final moles.

• Example Reaction: $\text{KBr} (aq) + \text{AgNO}_3 (aq) \rightarrow \text{KNO}_3 (aq) + \text{AgBr} (s)$

	KBr (aq)	AgNO3 (aq)	KNO3 (aq)	AgBr (s)
Initial	0.050 mol	0.050 mol	0	0
Change	-0.050 mol	-0.050 mol	+0.050 mol	+0.050 mol
Final	0	0	0.050 mol	0.050 mol

Empirical and Molecular Formulas

Determining Formulas from Experimental Data

Empirical formulas show the simplest whole-number ratio of elements in a compound, while molecular formulas show the actual number of atoms of each element in a molecule.

• Use percent composition and molar mass to determine the molecular formula.

• Example: Aspirin is composed of 60.0% C, 35.5% O, and the remainder H. If the molar mass is 150-200 g/mol, calculate the molecular formula.

Solution Stoichiometry and Dilutions

Calculating Concentrations and Volumes

Solution stoichiometry involves calculations with concentrations (molarity) and volumes to determine the amount of solute or product formed.

• Dilution Equation: $M_1V_1 = M_2V_2$ Where $M$ is molarity and $V$ is volume.

• Example: To prepare 250 mL of 0.100 M NaCl from a 1.00 M stock solution, use: $1.00 \times V_1 = 0.100 \times 0.250$ $V_1 = 0.025$ L = 25 mL

Summary Table: Precipitation Reaction Data

The following table summarizes the changes in moles for a precipitation reaction:

Step	KBr (aq)	AgNO3 (aq)	KNO3 (aq)	AgBr (s)
Initial	0.050 mol	0.050 mol	0	0
Change	-0.050 mol	-0.050 mol	+0.050 mol	+0.050 mol
Final	0	0	0.050 mol	0.050 mol

*Additional info: The tables and molecular views help visualize the stoichiometry and limiting reactant concepts in precipitation reactions.*