Chapter 4 Stoichiometry of Chemical Reactions

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Presentation transcript:

Chapter 4 Stoichiometry of Chemical Reactions Part A

Writing and Balancing Chemical Equations Two step process Write unbalanced equation Given products and reactants. All atoms present in reactants must also be present among products The physical states are indicated with (s) for solids, (l ) for liquids, (g ) for gases, and (aq) for aqueous solutions Adjust coefficients to get equal numbers of each kind of atom on both sides of arrow

Writing and Balancing Chemical Equations Reaction of methane with oxygen Write reactants on the left of the arrow: Methane CH4(g) and oxygen O2(g) 2. Write products on the right of the arrow: Carbon dioxide CO2(g) and water H2O(l ) 3. Adjust coefficients to get equal numbers of each kind of atom on both sides of arrow CH4(g) + 2O2(g)  CO2(g) + 2H2O(l ) One methane molecule and two oxygen molecules react to yield one carbon dioxide molecule and two water molecules.

Guidelines for Balancing Equations Start balancing with the most complicated formula first Balance atoms that appear in only two formulas: one as a reactant and the other as a product Leave elements that appear in three or more formulas until later

Reaction of propane with oxygen C3H8(g) + O2(g)  CO2(g) + H2O(l ) Assume 1 in front of C3H8 3 C 1 C  3 8 H 2 H  4 1C3H8(g) + __O2(g)  3CO2(g) + 4H2O(l ) O=2  5 = 10 O = (3  2) + 4 = 10 C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l )

Equations for Ionic Reactions Balance as a group those polyatomic ions that appear unchanged on both sides of the arrow AgNO3(aq) + Na3PO4(aq)  Ag3PO4(s) + NaNO3(aq) Adjust coefficients of both AgNO3 and NaNO3 to three. 3AgNO3(aq) + Na3PO4(aq)  Ag3PO4(s) + 3NaNO3(aq) Now check polyatomic ions 3 NO3– 3 NO3– 1 PO43– 1 PO43– This balanced equation is called a molecular equation or full formula equation.

Complete Ionic Equation Let consider the full formula equation CaCl2(aq) + 2AgNO3(aq)⟶Ca(NO3)2(aq) + 2AgCl(s) Ionic compounds dissolved in water are dissociated into ions CaCl2(aq)⟶Ca2+ (aq) + 2Cl−(aq) 2AgNO3(aq)⟶2Ag+(aq) + 2NO3−(aq) Ca(NO3)2(aq)⟶Ca2+ (aq) + 2NO3−(aq) AgCl does not dissolve in water In a complete ionic equation the formulas for the dissolved ionic compounds are replaced by their dissociated ions: Ca2+ (aq) + 2Cl−(aq) + 2Ag+(aq) + 2NO3−(aq) ⟶ Ca2+ (aq) + 2NO3−(aq) + 2AgCl(s)

Cl−(aq) + Ag+(aq)⟶AgCl(s) Net Ionic Equation Two chemical species Ca2+ (aq) and NO3−(aq) are present in identical form on both sides of the arrow. Ca2+ (aq) + 2Cl−(aq) + 2Ag+(aq) + 2NO3−(aq) ⟶ Ca2+ (aq) + 2NO3−(aq) + 2AgCl(s) These spectators may be eliminated from the equation to yield a net ionic equation: Cl−(aq) + Ag+(aq)⟶AgCl(s) The same number of each kind of atom must be present on both sides of the arrow. The net electrical charge on the reactants must equal the net electrical charge on the products Solid silver chloride may be produced from dissolved chloride and silver(I) ions, regardless of the source of these ions.

Classifying Chemical Reactions Double Replacement reactions Double replacement reaction involve the exchange of ions between ionic compounds. 1. Precipitation Reactions 2. Acid-Base Reactions 3. Gas Formation Reaction Oxidation-Reduction Reactions

Precipitation Reactions A precipitation reaction is one in which dissolved substances react to form solid products Example: the formation of solid lead iodide: 2KI(aq)+Pb(NO3)2(aq) ⟶PbI2(s)+2KNO3(aq) The net ionic equation representing this reaction Pb2+ (aq)+2I−(aq) ⟶ PbI2(s)

Solubility Rules Soluble compounds contain • group 1 metal cations , Li+, Na+, K+, Rb+, and Cs+ and ammonium ion NH4+ • the halide ions Cl−, Br−, I− • the acetate (C2H3O2−), bicarbonate (HCO3−), nitrate (NO3−), and chlorate (ClO3−) • the sulfate (SO42−) ion Exceptions to these solubility rules include • halides of Ag+, Hg22+ and Pb 2+ • sulfates of Ag+, Ba 2+ ,Ca 2+ ,Hg22+ , Pb2+, and Sr 2+ Insoluble compounds contain • carbonate (CO32−), chromate (CrO42−), phosphate (PO43−), and sulfide (S2−) ions • hydroxide ion (OH−) Exceptions to these insolubility rules • compounds of these anions with group 1 metal cations and ammonium ion • hydroxides of group 1 metal cations and Ba2+

Predicting Precipitation Reactions Predict the result of mixing of the following ionic compounds. (a) potassium sulfate and barium nitrate (b) lithium chloride and silver acetate (c) lead nitrate and ammonium carbonate

HCl(aq)+H2O(aq) ⟶ Cl−(aq)+H3O+(aq) Acid-Base Reactions An acid-base reaction is one in which a hydrogen ion, H+, is transferred from one chemical species to another. An acid is a substance that will dissolve in water to yield hydronium ions, H3O+ HCl(aq)+H2O(aq) ⟶ Cl−(aq)+H3O+(aq) HCl, HBr, HI, HNO3, HClO3, HClO4, H2SO4 react completely with water are called strong acids. Weak acids only partially react with water CH3CO2H(aq)+H2O(l)⇌CH3CO2−(aq)+H3O+(aq)

Base A base is a substance that will dissolve in water to yield hydroxide ions, OH−. NaOH, when dissolved in water is completely dissociated to Na+ and OH− ions, is called a strong base. NaOH(s) ⟶ Na+(aq)+OH−(aq) Ammonia which reacts partially with water to yield hydroxide ions, is a weak base. NH3(aq)+H2O(l)⇌NH4+(aq)+OH−(aq)

Formation of molecular compound An acid can react with a base to form water, which is a molecular compound.(neutralization reaction) HCl(aq) + NaOH(aq) ⟶ NaCl(aq) + H2O(l) Acid HCl can react with sodium acetate to form acetic acid, which exists mostly in molecular form. HCl(aq) + NaCH3COO(aq) ⟶ NaCl(aq) + HCH3COO(aq) Acid HCl can react with sodium cyanide to form the gas hydrogen cyanide. HCl(aq) + NaCN(aq) ⟶ NaCl(aq) + HCN(g)

Oxidation-Reduction Reactions Oxidation-reduction (redox) reactions involve the transfer of electrons between reactant species. 2Na(s)+Cl2(g) ⟶ 2NaCl(s) The reaction can be represented by two half-reactions 2Na(s) ⟶ 2Na+(s)+2e− oxidation = loss of electrons reducing agent = species that is oxidized Cl2(g)+2e− ⟶ 2Cl−(s) reduction = gain of electrons oxidizing agent = species that is reduced

Oxidation number Oxidation number (or oxidation state) of an element in a compound is the charge its atoms would possess if the compound was ionic. Guidelines to assign oxidation numbers The oxidation number of an atom in an elemental substance is zero. The oxidation number of a monatomic ion is equal to the ion’s charge. Oxidation numbers for common nonmetals are usually assigned as follows: • Hydrogen: +1 when combined with nonmetals, −1 when combined with metals

Oxidation number • Oxygen: −2 in most compounds • Halogens: −1 The sum of oxidation numbers for all atoms in a molecule or polyatomic ion equals the charge on the molecule or ion. New definition of Redox reaction oxidation = increase in oxidation number reduction = decrease in oxidation number Fe2+(aq)+Ag+(aq) ⟶ Fe3+ (aq)+Ag(s)

Oxidation reduction reaction Propellant reaction in which solid aluminum is oxidized by ammonium perchlorate 10Al(s)+6NH4ClO4(s) ⟶ 4Al2O3(s)+2AlCl3(s)+12H2O(g)+3N2(g) Single-displacement (replacement) reactions are redox reactions in which an ion in solution is displaced (or replaced) via the oxidation of a metallic element . Examples Zn(s)+2HCl(aq) ⟶ ZnCl2(aq)+H2(g) Cu(s)+2AgNO3(aq) ⟶ Cu(NO3)2(aq)+2Ag(s)

Balancing Redox Reactions via the Half-Reaction Method in Acidic Solution 1. Write the two half-reactions representing the redox process. 2. Balance all elements except oxygen and hydrogen. 3. Balance oxygen atoms by adding H2O molecules. 4. Balance hydrogen atoms by adding H+ ions. 5. Balance charge by adding electrons. 6. If necessary, multiply each half-reaction’s coefficients by the smallest possible integers to yield equal numbers of electrons in each. 7. Add the balanced half-reactions together and simplify by removing species that appear on both sides of the equation. 8. Finally, check to see that both the number of atoms and the total charges are balanced

Balancing Redox Reactions via the Half-Reaction Method in Basic Solution For reactions occurring in basic media (excess hydroxide ions), carry out these additional steps: a. Add OH− ions to both sides of the equation in numbers equal to the number of H+ ions. b. On the side of the equation containing both H+ and OH− ions, combine these ions to yield water molecules. c. Simplify the equation by removing any redundant water molecules Or use the same procedure as for acidic solution but balance oxygen atoms by using (O) + H2O = 2 OH- And balance hydrogen atoms by (H) + OH- = H2O