General Chemistry Exam Study Guide: Chemical Reactions, Stoichiometry, and Solution Chemistry

Chemical Equations and Balancing

Balancing Chemical Equations

Balancing chemical equations ensures that the same number of atoms of each element are present on both sides of the reaction. This is required by the Law of Conservation of Mass.

• Coefficients are placed in front of chemical formulas to indicate the number of molecules or moles involved.

• Balance elements one at a time, starting with the most complex molecule.

• Example: To balance $H_3PO_4(aq) + NaOH(aq) \rightarrow Na_3PO_4(aq) + H_2O(l)$, assign coefficients so that the number of each atom is equal on both sides.

Example: Balancing $C_4H_8(g) + O_2(g) \rightarrow CO_2(g) + H_2O(g)$ requires adjusting coefficients for carbon, hydrogen, and oxygen.

Types of Chemical Reactions

Classification of Reactions

Chemical reactions are classified based on the rearrangement of atoms and molecules.

• Synthesis (Combination): Two or more substances combine to form one product.

• Decomposition: A single compound breaks down into two or more simpler substances.

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

• Double Replacement: Exchange of ions between two compounds.

Example: $HClO_4(aq) + KOH(aq) \rightarrow KClO_4(aq) + H_2O(l)$ is a double-replacement reaction.

Electrolytes and Solubility

Strong Electrolytes

Strong electrolytes dissociate completely in water, producing ions that conduct electricity.

• Examples: NaCl, HCl, KOH

• Weak electrolytes: Partially dissociate (e.g., CH3COOH)

• Nonelectrolytes: Do not dissociate (e.g., C6H12O6)

Precipitation Reactions

Precipitation occurs when two aqueous solutions combine to form an insoluble solid (precipitate).

• Use solubility rules to predict precipitate formation.

• Example: Mixing $BaCl_2(aq)$ and $LiOH(aq)$ may form $Ba(OH)_2(s)$ if insoluble.

Oxidation-Reduction (Redox) Reactions

Identifying Redox Reactions

Redox reactions involve the transfer of electrons between species, changing their oxidation states.

• Oxidation: Loss of electrons (increase in oxidation state).

• Reduction: Gain of electrons (decrease in oxidation state).

• Example: $Na(s) + Cl_2(g) \rightarrow 2NaCl(s)$ (Na is oxidized, Cl is reduced).

Oxidation States

The oxidation state (number) indicates the degree of oxidation of an atom in a compound.

• Assign oxidation numbers based on rules (e.g., O is usually -2, H is +1).

• Chlorine can have various oxidation states, such as -1 in Cl-, +5 in ClO3-.

Stoichiometry and Calculations

Mole Calculations

Stoichiometry uses balanced equations to relate quantities of reactants and products.

• Mole ratio: Derived from coefficients in the balanced equation.

• Example: $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$ (1 mol N2 reacts with 3 mol H2 to produce 2 mol NH3).

Percent Yield

Percent yield measures the efficiency of a reaction.

• $\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$

• Example: If theoretical yield is 3.40 mol and actual yield is 3.22 mol, percent yield is $\frac{3.22}{3.40} \times 100\% = 94.7\%$.

Limiting Reactant and Mass Calculations

The limiting reactant is the reactant that is completely consumed first, limiting the amount of product formed.

• Calculate moles of each reactant and compare using the balanced equation.

• Use molar mass to convert between grams and moles.

Solution Concentration and Neutralization

Concentration is often expressed in molarity ($M$), defined as moles of solute per liter of solution.

• $M = \frac{\text{moles of solute}}{\text{liters of solution}}$

• Neutralization reactions involve acid and base reacting to form water and a salt.

• Example: To neutralize $1.25\,L$ of $0.255\,M$ $H_3PO_4$ with NaOH, use stoichiometry to find required mass of NaOH.

Combustion Analysis and Empirical/Molecular Formulas

Combustion Analysis

Combustion analysis determines the empirical formula of an organic compound by measuring the amounts of CO2 and H2O produced.

• Calculate moles of C and H from CO2 and H2O masses.

• Subtract C and H masses from total to find O mass.

• Use molar mass to determine molecular formula.

Solubility Rules and Activity Series

Solubility Table

Solubility rules help predict whether a compound will dissolve in water or form a precipitate.

Ion	NO3-	Cl-, Br-, I-	CO32-, SO32-, PO43-, CrO42-	SO42-	S2-	OH-
Li+, Na+, K+, NH4+	s	s	s	s	ss	s
Mg2+	s	s	i	s	ss	i
Ca2+, Sr2+, Ba2+	s	s	i	ss	ss	ss
Ag+	s	i	i	i	i	i
Pb2+	s	i	i	i	i	i
Fe2+, Zn2+, Cu2+	s	s	i	i	i	i

s = soluble, i = insoluble, ss = slightly soluble, d = decomposes in water

Activity Series

The activity series ranks metals by their ability to displace other metals in single replacement reactions.

• Metals higher in the series are more reactive.

• Example: Li > K > Ba > Ca > Na > Mg > Al > Mn > Zn > Cr > Fe > Ni > Sn > Pb > H2 > Cu > Ag > Au

Sample Problems and Applications

Stoichiometry and Limiting Reactant

Given masses of reactants, determine the limiting reactant and calculate the mass of product formed.

• Convert grams to moles using molar mass.

• Use mole ratios from the balanced equation.

Combustion Analysis

Calculate the molecular formula of an unknown compound from combustion data.

• Find moles of C and H from CO2 and H2O produced.

• Subtract from total mass to find O.

• Divide by smallest number of moles to get empirical formula.

• Use molar mass to find molecular formula.

Fortified Wine Calculation

Calculate volumes needed to mix solutions of different concentrations to achieve a desired final concentration.

• Use the dilution equation: $C_1V_1 + C_2V_2 = C_fV_f$

• Solve for unknown volumes.

Redox Reaction Example: Thermite Reaction

The thermite reaction involves aluminum powder and iron(III) oxide to produce iron metal and aluminum oxide.

• Type: Redox, single replacement

• Balanced equation: $2Al(s) + Fe_2O_3(s) \rightarrow 2Fe(l) + Al_2O_3(s)$

• Calculate grams of iron produced using stoichiometry.

Additional info: Some explanations and examples have been expanded for clarity and completeness.