Chemical Reactions and Stoichiometry: Study Guide

Module 4: Chemical Reactions

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in general chemistry, ensuring that the law of conservation of mass is obeyed. Each chemical equation must have the same number of atoms of each element on both sides.

• Chemical Equation Structure: Reactants are written on the left, products on the right, separated by an arrow (→).

• Symbols and Abbreviations:

  • (s): Solid

  • (l): Liquid

  • (g): Gas

  • (aq): Aqueous (dissolved in water)

  • Δ: Energy or heat

• Stoichiometric Coefficient: The number before a compound, indicating the number of molecules or moles.

• Subscripts: Indicate the number of atoms in a molecule.

Example: $6\ CO_2\ (g) + 6\ H_2O\ (l) + \Delta \rightarrow C_6H_{12}O_6\ (aq) + 6\ O_2\ (g)$

Steps to Balance a Chemical Equation

1. Identify the most complex substance.

2. Choose an element that appears in only one reactant and one product; adjust coefficients to balance.

3. Balance polyatomic ions as a unit if present on both sides.

4. Balance remaining atoms, ending with the least complex substance. Use fractional coefficients if needed, then multiply to obtain whole numbers.

5. Count atoms on both sides to confirm balance.

Example: Combustion of heptane: $C_7H_{16}\ (l) + 11\ O_2\ (g) \rightarrow 7\ CO_2\ (g) + 8\ H_2O\ (g)$

General Types of Chemical Reactions

Classification of Chemical Reactions

Chemical reactions can be classified into several types based on the nature of the reactants and products.

• Combustion: A substance reacts with oxygen to produce energy, often forming CO2 and H2O.

  • Example: $C_3H_8\ (g) + 5\ O_2\ (g) \rightarrow 3\ CO_2\ (g) + 4\ H_2O\ (g)$

• Combination (Synthesis): Two or more substances combine to form a new compound.

  • Example: $2\ Na\ (s) + Cl_2\ (g) \rightarrow 2\ NaCl\ (s)$

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

  • Example: $2\ HgO\ (s) \rightarrow 2\ Hg\ (l) + O_2\ (g)$

• Single Displacement: One element replaces another in a compound.

  • Example: $Mg\ (s) + Cu(NO_3)_2\ (aq) \rightarrow Mg(NO_3)_2\ (aq) + Cu\ (s)$

• Double Displacement: Ions in two compounds exchange places.

  • Example: $2\ KI\ (aq) + Pb(NO_3)_2\ (aq) \rightarrow 2\ KNO_3\ (aq) + PbI_2\ (s)$

• Acid-Base (Neutralization): Acid reacts with base to form salt and water.

  • Example: $HCl\ (aq) + NaOH\ (aq) \rightarrow NaCl\ (aq) + H_2O\ (l)$

Stoichiometry

Mole Ratios and Quantitative Relationships

Stoichiometry is the calculation of reactant and product quantities in chemical reactions, based on balanced equations.

• Mole Ratio: The ratio of moles of one substance to another, derived from the coefficients in the balanced equation.

• Example: In $C_3H_8\ (g) + 5\ O_2\ (g) \rightarrow 3\ CO_2\ (g) + 4\ H_2O\ (g)$, the mole ratio of propane to oxygen is 1:5.

Conversions

• Mol to Mol: Use mole ratios to convert between substances.

• Gram to Mol: Use molar mass ($\text{grams} \div \text{molar mass} = \text{moles}$).

• Mol to Gram: Use molar mass ($\text{moles} \times \text{molar mass} = \text{grams}$).

Example: If 3.5 moles of propane react, moles of oxygen required: $3.5\ \text{mol}\ \times \frac{5\ \text{mol}\ O_2}{1\ \text{mol}\ C_3H_8} = 17.5\ \text{mol}\ O_2$

The Limiting Reactant and Percent Yield

Limiting Reactant

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

• Identify Limiting Reactant: Compare the mole ratios of reactants to the balanced equation.

• Example: If 100 g of hydrogen and 100 g of oxygen react, calculate moles and determine which is limiting.

Theoretical and Actual Yield

Theoretical yield is the maximum amount of product that can be formed from the limiting reactant. Actual yield is the amount actually obtained from the experiment.

• Percent Yield Formula:

$\text{Percent Yield} = \frac{\text{Actual Yield (moles)}}{\text{Theoretical Yield (moles)}} \times 100\%$

• Example: If theoretical yield is 120 g and actual yield is 100 g, percent yield is $\frac{100}{120} \times 100\% = 83.3\%$

Sample Problem: Zinc and Silver Nitrate

• Given: 2.00 g Zn and 2.50 g AgNO3

• Find: Limiting reactant, grams of Ag formed, grams of Zn(NO3)2 formed, excess reactant left, percent yield if 1.0 g Ag recovered.

Summary Table: Types of Chemical Reactions

Type	General Equation	Example
Combustion	Substance + O2 → Products + Energy	C3H8 + 5 O2 → 3 CO2 + 4 H2O
Combination	A + B → AB	2 Na + Cl2 → 2 NaCl
Decomposition	AB → A + B	2 HgO → 2 Hg + O2
Single Displacement	A + BC → AC + B	Mg + Cu(NO3)2 → Mg(NO3)2 + Cu
Double Displacement	AB + CD → AD + CB	2 KI + Pb(NO3)2 → 2 KNO3 + PbI2
Acid-Base	Acid + Base → Salt + Water	HCl + NaOH → NaCl + H2O

Relevant Images

The following images are included as they directly relate to the classification and balancing of chemical reactions, visually reinforcing the types and steps discussed above.