Chapter 4: Chemical Reactions and Quantities - Detailed Notes

Chemical Reactions

• A chemical reaction is a process where one or more substances transform into one or more different substances.
• It involves chemical changes in matter, leading to the creation of new chemical substances.
• Reactants are the starting materials.
• Products are the newly formed substances.

Chemical Equations

• A chemical equation is a shorthand method for describing a chemical reaction.
• It uses the symbols ® to show the transformation from reactants to products:
  Reactants ® Products
• Chemical equations provide key information about the reaction:
  • Formulas of reactants and products
  • States of reactants and products (e.g., gas, liquid, solid, aqueous)
  • Relative numbers of reactant and product molecules. This information can be used to determine the masses of reactants used and products formed.

States of Reactants and Products

• The states of reactants and products are indicated using abbreviations:
  • (g): Gas
  • (l): Liquid
  • (s): Solid
  • (aq): Aqueous (water solution)

Balancing Chemical Equations

• Chemical equations must be balanced to ensure that the number of atoms of each element is the same on both sides of the reaction arrow.
• Balancing equations is essential for consistency with the Law of Conservation of Mass.
• The coefficients in a balanced chemical equation allow us to predict the relative amounts of reactants and products.
• Chemical equations can be interpreted in terms of moles of reactants and products.

Combustion Reactions

• A combustion reaction is a type of chemical reaction where a substance combines with oxygen to form one or more oxygen-containing compounds.
• Combustion reactions also release heat.

Combustion of Methane

• Methane gas (CH$_{4}$) burns in the presence of oxygen to produce carbon dioxide gas and gaseous water.
• The unbalanced equation is: $CH<em>4(g) + O</em>2(g) \rightarrow CO<em>2(g) + H</em>2O(g)$
• To balance the equation, adjust the coefficients to ensure equal numbers of atoms on both sides.

Balancing Chemical Equations: Example 4.1

• Problem: Write a balanced equation for the reaction between solid cobalt(III) oxide ($Co<em>2O</em>3$) and solid carbon (C) to produce solid cobalt (Co) and carbon dioxide gas ($CO_2$).
  1. Write the skeletal equation: $Co<em>2O</em>3(s) + C(s) \rightarrow Co(s) + CO_2(g)$
  2. Balance atoms in complex substances first: Start with oxygen.
     • $Co<em>2O</em>3(s) + C(s) \rightarrow Co(s) + CO_2(g)$
     • 3 O atoms on the left, 2 O atoms on the right
     • To balance O, add coefficients: $2Co<em>2O</em>3(s) + C(s) \rightarrow Co(s) + 3CO_2(g)$
     • Now there are 6 O atoms on each side.
  3. Balance free elements: Balance cobalt next.
     • $2Co<em>2O</em>3(s) + C(s) \rightarrow Co(s) + 3CO_2(g)$
     • 4 Co atoms on the left, 1 Co atom on the right
     • To balance Co, add a coefficient: $2Co<em>2O</em>3(s) + C(s) \rightarrow 4Co(s) + 3CO_2(g)$
     • Now there are 4 Co atoms on each side.
  4. Balance remaining elements: Balance carbon.
     • $2Co<em>2O</em>3(s) + C(s) \rightarrow 4Co(s) + 3CO_2(g)$
     • 1 C atom on the left, 3 C atoms on the right
     • To balance C, add a coefficient: $2Co<em>2O</em>3(s) + 3C(s) \rightarrow 4Co(s) + 3CO_2(g)$
     • Now there are 3 C atoms on each side.
  5. Check the balance: Ensure the equation is balanced by summing the total number of each type of atom on both sides.
     • $2Co<em>2O</em>3(s) + 3C(s) \rightarrow 4Co(s) + 3CO_2(g)$
     • Left: 4 Co atoms, 6 O atoms, 3 C atoms
     • Right: 4 Co atoms, 6 O atoms, 3 C atoms
     • The equation is balanced.

Chemical Equations Containing Polyatomic Ions: Example 4.3

• Problem: Write a balanced equation for the reaction between aqueous strontium chloride ($SrCl<em>2$) and aqueous lithium phosphate ($Li</em>3PO<em>4$) to form solid strontium phosphate ($Sr</em>3(PO<em>4)</em>2$) and aqueous lithium chloride (LiCl).
  1. Write the skeletal equation: $SrCl<em>2(aq) + Li</em>3PO<em>4(aq) \rightarrow Sr</em>3(PO<em>4)</em>2(s) + LiCl(aq)$
  2. Balance metal ions (cations) first: Start with strontium ($Sr^{2+}$).
     • $SrCl<em>2(aq) + Li</em>3PO<em>4(aq) \rightarrow Sr</em>3(PO<em>4)</em>2(s) + LiCl(aq)$
     • 1 $Sr^{2+}$ ion on the left, 3 $Sr^{2+}$ ions on the right
     • To balance, add coefficient: $3SrCl<em>2(aq) + Li</em>3PO<em>4(aq) \rightarrow Sr</em>3(PO<em>4)</em>2(s) + LiCl(aq)$
     • Now there are 3 $Sr^{2+}$ ions on each side.
  3. Balance lithium ions ($Li^+$): $3SrCl<em>2(aq) + Li</em>3PO<em>4(aq) \rightarrow Sr</em>3(PO<em>4)</em>2(s) + LiCl(aq)$
     • 3 $Li^+$ ions on the left, 1 $Li^+$ ion on the right
     • To balance, add coefficient: $3SrCl<em>2(aq) + Li</em>3PO<em>4(aq) \rightarrow Sr</em>3(PO<em>4)</em>2(s) + 3LiCl(aq)$
     • Now there are 3 $Li^+$ ions on each side.
  4. Balance nonmetal ions (anions) second: Start with phosphate ($PO_4^{3-}$).
     • $3SrCl<em>2(aq) + Li</em>3PO<em>4(aq) \rightarrow Sr</em>3(PO<em>4)</em>2(s) + 3LiCl(aq)$
     • 1 $PO<em>4^{3-}$ ion on the left, 2 $PO</em>4^{3-}$ ions on the right
     • To balance, add coefficient: $3SrCl<em>2(aq) + 2Li</em>3PO<em>4(aq) \rightarrow Sr</em>3(PO<em>4)</em>2(s) + 3LiCl(aq)$
     • Now there are 2 $PO_4^{3-}$ ions on each side.
  5. Balance chloride ions ($Cl^-$): Since adjusting PO4 threw off the Li balance, we must revisit it.
     • $3SrCl<em>2(aq) + 2Li</em>3PO<em>4(aq) \rightarrow Sr</em>3(PO<em>4)</em>2(s) + 3LiCl(aq)$
     • 6 $Cl^-$ ions on the left, 3 $Cl^-$ ions on the right
     • To balance, change coefficient: $3SrCl<em>2(aq) + 2Li</em>3PO<em>4(aq) \rightarrow Sr</em>3(PO<em>4)</em>2(s) + 6LiCl(aq)$
     • Now there are 6 $Cl^-$ ions on each side.
  6. Recheck Lithium Balance: We also need to update the lithium balance
     • $3SrCl<em>2(aq) + 2Li</em>3PO<em>4(aq) \rightarrow Sr</em>3(PO<em>4)</em>2(s) + 6LiCl(aq)$
     • 6 Li ions on the left, 6 Li ions on the right
  7. Check the balance: Ensure the equation is balanced by summing the total number of each type of ion on both sides.
     • $3SrCl<em>2(aq) + 2Li</em>3PO<em>4(aq) \rightarrow Sr</em>3(PO<em>4)</em>2(s) + 6LiCl(aq)$
     • Left: 3 $Sr^{2+}$ ions, 6 $Li^+$ ions, 2 $PO_4^{3-}$ ions, 6 $Cl^-$ ions
     • Right: 3 $Sr^{2+}$ ions, 6 $Li^+$ ions, 2 $PO_4^{3-}$ ions, 6 $Cl^-$ ions
     • The equation is balanced.

Conceptual Connection 4.1

• Question: How many oxygen atoms are on the right-hand side of the chemical equation $4FeCO<em>3(s) + O</em>2(g) \rightarrow 2Fe<em>2O</em>3(s) + 4CO_2(g)$?
• Answer: d. 14

Conceptual Connection 4.2

• Question: Which quantity or quantities must always be the same on both sides of a chemical equation?
• Answer: a. the number of atoms of each kind

Quantities in Chemical Reactions

• The amount of every substance used and made in a chemical reaction is related to the amounts of all other substances in the reaction.
• This is based on the Law of Conservation of Mass, which is enforced by balancing chemical equations.
• Stoichiometry is the study of the numerical relationships between chemical quantities in a chemical reaction.

Reaction Stoichiometry

• The coefficients in a chemical equation specify the relative amounts in moles of each of the substances involved in the reaction.
• Example: $2C<em>8H</em>{18}(l) + 25O<em>2(g) \rightarrow 16CO</em>2(g) + 18H_2O(g)$
  • 2 molecules of $C<em>8H</em>{18}$ react with 25 molecules of $O<em>2$ to form 16 molecules of $CO</em>2$ and 18 molecules of $H_2O$.
  • 2 moles of $C<em>8H</em>{18}$ react with 25 moles of $O<em>2$ to form 16 moles of $CO</em>2$ and 18 moles of $H_2O$.
  • Mole ratios: 2 mol $C<em>8H</em>{18}$ : 25 mol $O<em>2$ : 16 mol $CO</em>2$ : 18 mol $H_2O$
• Mole-to-mole ratios can be written from stoichiometric coefficients, e.g., $\frac{2 \text{ moles } C<em>8H</em>{18}}{25 \text{ moles } O<em>2}$ or $\frac{16 \text{ moles } CO</em>2}{18 \text{ moles } H_2O}$

Mole-to-Mole Conversions

• Example: If 22.0 moles of $C<em>8H</em>{18}$ are burned, how many moles of $CO_2$ form?
• The balanced equation is: $2C<em>8H</em>{18}(l) + 25O<em>2(g) \rightarrow 16CO</em>2(g) + 18H_2O(g)$
• Stoichiometric ratio: 2 moles $C<em>8H</em>{18}$ : 16 moles $CO_2$
• Calculation: $22.0 \text{ mol } C<em>8H</em>{18} \times \frac{16 \text{ mol } CO<em>2}{2 \text{ mol } C</em>8H<em>{18}} = 176 \text{ mol } CO</em>2$
• The combustion of 22.0 moles of $C<em>8H</em>{18}$ produces 176 moles of $CO_2$.

Making Pizza Analogy

• Relating chemical reactions to a real-world example.
• Example: 1 crust + 5 oz tomato sauce + 2 cups cheese → 1 pizza
• The relationship can be expressed mathematically: 1 crust : 5 oz sauce : 2 cups cheese : 1 pizza
• If you have 6 cups of cheese: $\frac{6 \text{ cups cheese}}{1} \times \frac{1 \text{ pizza}}{2 \text{ cups cheese}} = 3 \text{ pizzas}$

Making Molecules: Mole-to-Mole Conversions

• Using the stoichiometric ratio from the balanced chemical equation as a conversion factor.
• Example: $2C<em>8H</em>{18}(l) + 25O<em>2(g) \rightarrow 16CO</em>2(g) + 18H_2O(g)$
• Stoichiometric ratio: 2 moles $C<em>8H</em>{18}$ : 16 moles $CO_2$

Conceptual Connection 4.3

• Question: Use the balanced equation $2C<em>8H</em>{18}(l) + 25O<em>2(g) \rightarrow 16CO</em>2(g) + 18H<em>2O(g)$ to determine how many moles of $H</em>2O$ are produced by the combustion of 22.0 moles of $C<em>8H</em>{18}$.
• Answer: d. 198 moles $H_2O$

Making Molecules: Mass-to-Mass Conversions

• Using molar mass and coefficients as conversion factors.
• Molar mass is used as a conversion factor between mass and amount in moles.
• Coefficients are used as the conversion factor between the amount in moles of reactants and products.

Stoichiometry: Example 4.4

• Photosynthesis reaction: $6CO<em>2(g) + 6H</em>2O(l) \xrightarrow{\text{sunlight}} C<em>6H</em>{12}O<em>6(aq) + 6O</em>2(g)$
• Problem: If a plant consumes 37.8 g of $CO<em>2$ in one week, what mass of glucose ($C</em>6H<em>{12}O</em>6$) can it synthesize?
  1. Sort: Given: 37.8 g $CO<em>2$, Find: g $C</em>6H<em>{12}O</em>6$
  2. Strategize: mass A → amount A (in moles) → amount B (in moles) → mass B
  3. Conceptual Plan:
     • Molar mass $CO_2$ = 44.01 g/mol
     • 6 mol $CO<em>2$ : 1 mol $C</em>6H<em>{12}O</em>6$
     • Molar mass $C<em>6H</em>{12}O_6$ = 180.2 g/mol
  4. Solve:$\frac{37.8 \text{ g } CO<em>2}{1} \times \frac{1 \text{ mol } CO</em>2}{44.01 \text{ g } CO<em>2} \times \frac{1 \text{ mol } C</em>6H<em>{12}O</em>6}{6 \text{ mol } CO<em>2} \times \frac{180.2 \text{ g } C</em>6H<em>{12}O</em>6}{1 \text{ mol } C<em>6H</em>{12}O<em>6} = 25.8 \text{ g } C</em>6H<em>{12}O</em>6$

Conceptual Connection 4.4

• Reaction: $4Na(s) + O<em>2(g) \rightarrow 2Na</em>2O(s)$
• Question: Which image best represents the amount of sodium required to completely react with all of the oxygen?

Limiting Reactant, Theoretical Yield, Percent Yield

• Limiting Reactant: The reactant that is completely consumed first and limits the amount of product formed.
• Theoretical Yield: The maximum amount of product that can be made based on the amount of limiting reactant.
• Percent Yield: The actual yield (amount of product actually obtained) expressed as a percentage of the theoretical yield.

Percent Yield Formula

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

Key Definitions

• Limiting Reactant: The reactant that is completely consumed in a chemical reaction and limits the amount of product formed.
• Reactant in Excess: Any reactant present in a quantity greater than is required to completely react with the limiting reactant.
• Theoretical Yield: The amount of product that can be made in a chemical reaction based on the amount of limiting reactant.
• Actual Yield: The amount of product actually produced by a chemical reaction, determined experimentally.
• Percent Yield: The actual yield divided by the theoretical yield, multiplied by 100%.

Calculating Limiting Reactant

• Convert reactant masses to moles.
• Use stoichiometry to determine which reactant will produce less product.

Example - Limiting Reactant

• 2 Mg + O2 → 2 MgO
• A reactant mixture contains 42.5 g Mg and 33.8 g O2.

Conceptual Connection 4.7

• Question: $3NO<em>2(g) + H</em>2O(l) \rightarrow 2HNO_3(l) + NO(g)$
• Suppose that 5 mol $NO<em>2$ and 1 mol $H</em>2O$ combine and react completely. How many moles of the reactant in excess are present after the reaction has completed?
• Answer: c. 2 mol $NO_2$

Limiting Reactant and Theoretical Yield: Example 4.6

• Reaction: $2NO(g) + 5H<em>2(g) \rightarrow 2NH</em>3(g) + 2H_2O(g)$
• Starting with 86.3 g NO and 25.6 g H2, find the theoretical yield of ammonia in grams.

Actual Yield and Percent Yield

• The actual yield is the amount of product made in a chemical reaction and is determined experimentally.
• Percent yield = actual yield/ theoretical yield X 100%

Combustion Reactions

• Reactions with $O_2$ to form one or more oxygen-containing compounds, often including water.
• Example: $CH<em>4(g) + 2O</em>2(g) \rightarrow CO<em>2(g) + 2H</em>2O(g)$
• Ethanol combustion Example: $C<em>2H</em>5OH(l) + 3O<em>2(g) \rightarrow 2CO</em>2(g) + 3H_2O(g)$

Alkali Metal Reactions

• Reactions of alkali metals with nonmetals are vigorous.
• Sodium and chlorine form sodium chloride: $2Na(s) + Cl_2(g) \rightarrow 2NaCl(s)$
• Alkali metals with water: $2M(s) + 2H<em>2O(l) \rightarrow 2M^+(aq) + 2OH^-(aq) + H</em>2(g)$

Halogen Reactions

• Halogens react with many metals to form metal halides.
• $2Fe(s) + 3Cl<em>2(g) \rightarrow 2FeCl</em>3(s)$
• Halogens react with hydrogen to form hydrogen halides.
• $H<em>2(g) + X</em>2(g) \rightarrow 2HX(g)$
• Halogens react with each other to form interhalogen compounds.
• $Br<em>2(l) + F</em>2(g) \rightarrow 2BrF(g)$