By Professor Walker Waugh Balancing Reactions By Professor Walker Waugh

LAW OF CONSERVATION OF MASS Balancing Reactions Remember the LAW OF CONSERVATION OF MASS Matter CANNOT be created nor destroyed! So USE COMMON SENSE: However many atoms you have on the reactant side you must also have on the product side

Method for Balancing Reactions Treat all polyatomic ions as ONE UNIT, if it does not change from reactant to product Balance all NON-hydrogen and oxygen atoms Balance all hydrogen atoms Balance all oxygen atoms Check yourself before you WRECK yourself!

Simple things to watch for when Balancing Reactions Remember you can change the coefficients but NOT the subscripts Combustion reactions means that when a substance is burned it reacts with O2 (g) Oxygen MUST be present for something to burn, it supports combustion! Odd number of oxygen atoms on the product side Oxygen atoms in the reactant being burned

Balancing Pitfall #1 Odd number of oxygen atoms on the product side Butane gas (C4H10) burns C4H10 (g) + O2 (g)  CO2 (g) + H2O (g) NOT H2O (l) 1. Balance NON-hydrogen & oxygen atoms C4H10 (g) + O2 (g)  4 CO2 (g) + H2O (g) four C four C 2. Balance hydrogen atoms C4H10 (g) + O2 (g)  4 CO2 (g) + 5 H2O (g) ten H 10 H (because the coefficient 5 times the subscript 2 = 10) 3. Balance oxygen atoms two O eight O + five O = thirteen O 4. Make a fraction because 13/2 x 2 = 13 C4H10 (g) + 13/2 O2 (g)  4 CO2 (g) + 5 H2O (g)

Balancing Pitfall #1 Final Answer: Multiply the ENTIRE equation by 2 to remove fraction (we don’t have partial compounds) 2 (C4H10 (g) + 13/2 O2 (g)  4 CO2 (g) + 5 H2O (g)) Final Answer: 2 C4H10 (g) + 13 O2 (g)  8 CO2 (g) + 10 H2O (g) Check It 8 C 8 20 H 20 26 O 16 + 20 = 26

Balancing Pitfall #2 Oxygen atoms in the burning reactant CH3OH (g) + O2 (g)  CO2 (g) + H2O (g) Same rules as before but subtract an oxygen atom from the reactant side before you try to balance O2 CH3OH (g) + 3/2 O2 (g)  CO2 (g) + 2H2O (g) Again: Multiply by 2 to remove the fraction

2 CH3OH (g) + 3 O2 (g)  2 CO2 (g) + 4 H2O (g) Balancing Pitfall #2 2( CH3OH (g) + 3/2 O2 (g)  CO2 (g) + 2H2O (g)) Final Answer: 2 CH3OH (g) + 3 O2 (g)  2 CO2 (g) + 4 H2O (g) Check It 2 C 2 8 H 8 2 + 6 = 8 O 4 + 4 = 8

By Professor Walker Waugh Double Displacement Reactions Precipitation, Neutralization & Gaseous Reactions By Professor Walker Waugh

Predicting Products: Double Displacement Reactions There are three types of Double Displacement Reactions Precipitation Reactions Acid/Base Neutralizations Gaseous Reactions All occur via the same reaction mechanism: Swinger’s Mechanism A+B− + C+D−  A+D− + C+B− Remember to write the Cations (+) first then the Anions (-) in each compound!

Predicting Products: Precipitation Reactions Method: Recognize the reaction type and write the correct formulas of the reactants: Salt + Salt Salt + Acid Salt + Base Swing the cations. Check the formulas of the products, make sure that the nomenclature is correct. Balance the reaction. Check the Solubility Table to see if there is a solid formed. Salt: A classification of chemical compounds A salt is any aqueous ionic compound that is NOT an acid (e.g. HCl(aq) ), base (e.g. NaOH) or oxide (e.g. Li2O = Lithium oxide)

Predicting Products: Precipitation Reactions Example Aqueous barium nitrate is placed in a test tube containing aqueous lithium phosphate (I don’t give products for Precipitation Reactions) Ba(NO3)2 (aq) + Li3PO4 (aq)  BaPO4 ___ + LiNO3 ___ BaPO4 is NOT a correct formula Ba3(PO4)2 is the correct formula LiNO3 is the correct formula Re-examine the formula and balance it, including correct product formulas: 3 Ba(NO3)2 (aq) + 2 Li3PO4 (aq)  Ba3(PO4)2 ___ + 6 LiNO3 ___

Predicting Products: Precipitation Reactions Finally, check the Solubility Table, located on last page of Reaction Table 100, to assign the states to the products. If the table says that the substance is INSOLUBLE then it does NOT dissolve in water = (s) If the Table says that the substance in SOLUBLE then it DOES dissolve in water = (aq) Make sure to read the EXCEPTIONS! For Ba3(PO4)2 see Rule #5 For LiNO3 see Rule #1 Final Answer 3 Ba(NO3)2 (aq) + 2 Li3PO4 (aq)  Ba3(PO4)2 (s) + 6 LiNO3 (aq)

Solubility Table: know if taking General Chemistry next semester SOLUBILITY RULES FOR IONIC COMPOUNDS Compounds containing Group IA metals, ammonium, acetates and nitrates are all soluble. Most halides (Group 7A - chlorides etc.) are soluble. Exceptions include Ag+1, Pb+2, and Hg2+2 halides. Most sulfates are soluble. Exceptions include Ba+2, Sr+2, Ag+1, Pb+2, and Ca+2 sulfates. Most hydroxides insoluble. Exceptions include hydroxides of Group 1A metals, ammonium, Ca+2, Sr+2, and Ba+2. Most phosphates, carbonates, chromates, and sulfides are insoluble. Exceptions include those compounds containing Group 1A metals and ammonium. In addition, all acids are soluble!

Predicting Products: Acid/Base Neutralization Reactions Method: Recognize the reaction type and write the correct formulas of the reactants: Acid + Base (ONLY as Reactants) Swing the cations. When H+ + OH− combines then make H-OH (which is H2O (l) ). Check the formulas of the products, make sure that the nomenclature is correct. Refer to the Solubility Table and assign states of products. Balance the reaction. Check the Solubility Table to see if there is a solid formed.

Predicting Products: Acid/Base Neutralization Reactions Example Hydrobromic acid is placed in a test tube containing aqueous calcium hydroxide HBr (aq) + Ca(OH)2 (aq)  CaBr___ + H-OH___ HBr (aq) + Ca(OH)2 (aq)  CaBr2___ + H-OH___ HBr (aq) + Ca(OH)2 (aq)  CaBr2 (aq) + H-OH (l) 2 HBr (aq) + Ca(OH)2 (aq)  CaBr2 (aq) + 2 H-OH (l) It is easier to balance the reaction when water is written as as H-OH in acid/base neutralization reactions. Final Answer 2 HBr (aq) + Ca(OH)2 (aq)  CaBr2 (aq) + 2 H2O (l)

Predicting Products: Gaseous Reactions There are four types of Gaseous Displacement Reactions that you must know! Carbonate (CO32-, or HCO3-) + Acid Reactions Sulfites (SO32-, or HSO3-) + Acid Reactions Sulfide (S2-) + Acid Reactions Ammonium ion (NH4+) + Hydroxide (OH-) Reactions All occur via the same reaction mechanism: Swinger’s Mechanism A+B− + C+D−  A+D− + C+B−

Predicting Products: Gaseous Carbonate Reactions Method: Recognize the reaction type and write the correct formulas of the reactants: CO32- + Acid HCO3− + Acid Swing the cations. When H2CO3(aq), carbonic acid, is formed then it automatically dissociates to H2O (l) and CO2 (g) Check the formulas of the products, make sure that the nomenclature is correct. Balance the reaction. Check the Solubility Table to see if there is a solid formed.

Predicting Products: Gaseous Carbonate Reactions Example Hydrochloric acid is placed in a beaker containing aqueous sodium carbonate HCl (aq) + Na2CO3 (aq)  NaCl (aq) + HCO3 (aq) HCO3 (aq) is NOT a good formula. H2CO3 (aq) = carbonic acid HCl (aq) + Na2CO3 (aq)  NaCl (aq) + H2CO3 (aq) Balance the formula since all formulas are correct: 2 HCl (aq) + Na2CO3 (aq)  2 NaCl (aq) + H2CO3 (aq) Remember that H2CO3 (aq)  H2O (l) + CO2 (g) Final Answer 2 HCl (aq) + Na2CO3 (aq)  2 NaCl (aq) + H2O (l) + CO2 (g)

Predicting Products: Gaseous Sulfite Reactions Method: Recognize the reaction type and write the correct formulas of the reactants: SO32- + Acid HSO3− + Acid Swing the cations. When H2SO3(aq), sulfurous acid, is formed then it automatically dissociates to H2O (l) and SO2 (g) Check the formulas of the products, make sure that the nomenclature is correct. Balance the reaction. Check the Solubility Table to see if there is a solid formed.

Predicting Products: Gaseous Sulfite Reactions Example Nitric acid is placed in a beaker containing aqueous sodium hydrogen sulfite HNO3 (aq) + NaHSO3 (aq)  NaNO3 (aq) + H-HSO3 (aq) H-HSO3 (aq) is really H2SO3 (aq) = sulfurous acid. Remember that H2SO3 (aq)  H2O (l) + SO2 (g) Balance the formula since all formulas are correct and check the Solubility Table for state of the products: Final Answer HNO3 (aq) + NaHSO3 (aq)  NaNO3 (aq) + H2O (l) + SO2 (g)

Predicting Products: Gaseous Sulfide Reactions Method: Recognize the reaction type and write the correct formulas of the reactants: S2- + Acid Swing the cations. When H2S results it is also a gaseous product. Check the formulas of the products, make sure that the nomenclature is correct. Balance the reaction. Check the Solubility Table to see if there is a solid formed.

Predicting Products: Gaseous Sulfide Reactions Example Hydrochloric acid is placed in a beaker containing aqueous lithium sulfide HCl (aq) + Li2S (aq)  LiCl___ + HS___ HS is NOT a good formula. H2S is hydrogen sulfide. Remember that H2S when formed is GASEOUS and smells like ROTTEN EGGS! Balance the formula since all formulas are correct and check the Solubility Table for state of the products: 2 HCl (aq) + Li2S (aq)  2 LiCl (aq) + H2S (g) Final Answer Same as Above

Predicting Products: Gaseous Ammonium Reactions Method: Recognize the reaction type and write the correct formulas of the reactants: NH4+ + OH− Swing the cations. When NH4OH (aq), ammonium hydroxide, is formed then it automatically dissociates to H2O (l) and NH3 (g) Check the formulas of the products, make sure that the nomenclature is correct. Balance the reaction. Check the Solubility Table to see if there is a solid formed.

Predicting Products: Gaseous Ammonium Reactions Example Aqueous ammonium chloride is placed in a beaker containing aqueous barium hydroxide NH4Cl (aq) + Ba(OH)2 (aq)  BaCl (aq) + NH4OH (aq) BaCl (aq) is NOT the correct formula for barium chloride. Barium chloride (aq) = BaCl2. Thus: 2 NH4Cl (aq) + Ba(OH)2 (aq)  BaCl2 (aq) + 2 NH4OH (aq) Remember that NH4OH = NH3 (g) + H2O (l) Final Answer 2 NH4Cl (aq) + Ba(OH)2 (aq)  BaCl2 (aq) + 2 NH3 (g) + 2 H2O (l)

Molecular, Ionic and Net Ionic Equations By Professor Walker Waugh

Molecular, Ionic and Net Ionic Equations Molecular Equations: We will write all compounds as molecules even if ionic Complete Ionic Equations (CIE): Write all ionic compounds and strong acids as ions. Do not break up precipitants or weak acids on product side. Know the 7 strong acids (any thing else is considered a weak acid) HCl (aq), HBr (aq), HI (aq), HNO3 (aq), H2SO4 (aq), HClO4 (aq), HClO3 (aq) (chloric acid is NOT noted in your textbook) Net Ionic Equations (NIE): Complete Ionic Equations – Spectator Ions Spectator Ions = any ion that does NOT change in any way from reactant to product side of the equation

Molecular, Ionic and Net Ionic Equations Molecular Equations = A Balanced Chemical Equation Example #1 Sulfuric acid is placed is an Erlenmeyer flask containing aqueous sodium hydroxide H2SO4 (aq) + 2 NaOH (aq)  Na2SO4 (aq) + 2 H2O (l) Follow the same method as before!

Molecular, Ionic and Net Ionic Equations Complete Ionic Equations = CIE Example #1 Sulfuric acid is placed is an Erlenmeyer flask containing aqueous sodium hydroxide 2 H+ (aq) + 2 SO42- (aq) + 2 Na+ (aq) + 2 OH- (aq)  2 Na+ (aq) + SO42_ (aq) + 2 H2O (l) Remember to depict: COEFFICIENT, CHARGE and STATE for each ion! Spectator Ions = Na+ and SO42_

Molecular, Ionic and Net Ionic Equations Net Ionic Equations = NIE Example #1 Sulfuric acid is placed is an Erlenmeyer flask containing aqueous sodium hydroxide Thus: 2 H+ (aq) + 2 OH− (aq)  2 H2O (l) But the final equation must have LOWEST COEFFICIENT RATIO Final Answer H+ (aq) + OH_ (aq)  H2O (l)

Molecular, Ionic and Net Ionic Equations Molecular Equations Example #2 Solutions of plumbous nitrate and lithium bromide are mixed in a test tube Pb(NO3)2 (aq) + 2 LiBr (aq)  PbBr2 (s) + 2 LiNO3 (aq) Remember that a precipitant does NOT dissolve in water, therefore it DOES NOT DISSOCIATE.

Molecular, Ionic and Net Ionic Equations Complete Ionic Equations Example #2 Solutions of plumbous nitrate and lithium bromide are mixed in a test tube Pb+ (aq) + 2 NO3− (aq) + 2 Li+ + 2 Br− (aq)  PbBr2 (s) + 2 Li+ (aq) + 2 NO3− (aq) Spectator Ions = Li+ and NO3−

Molecular, Ionic and Net Ionic Equations Example #2 Solutions of plumbous nitrate and lithium bromide are mixed in a test tube Pb+ (aq) + 2 Br− (aq)  PbBr2 (s) Final Answer Pb+ (aq) + 2 Br−  PbBr2 (s)

Single Displacement Reactions and Oxidation Numbers By Professor Walker Waugh

Single Displacement Reactions One type of Oxidation Reduction (Redox) Reactions Occurs via a Transfer of Electrons and you must assign the OXIDATION NUMBERS to each element. But 1st let’s discuss single displacement reactions. General Equation: A + BC  B + AC Metal Look at Activity Series

Single Displacement Reactions ACTIVITY SERIES FOR METALS (and HYDROGEN) Li highest activity  K Ca Na If A is higher than B then YES there will be a reaction Mg Al Zn  Zn+2 Cr  Cr+3 Fe  Fe+2  Cd  Cd+2 Ni  Ni+2 Sn  Sn+2 Pb  Pb+2   H2 Cu  Cu+2 If A is NOT higher than B then NO REACTION Ag  Ag+1 Hg  Hg+2 Au  Au+3 lowest activity

Single Displacement Reactions Example #1 A solution of silver nitrate is placed in a test tube containing copper metal Cu (s) + 2 AgNO3 (aq)  Cu(NO3)2 (aq) + 2 Ag (s) A B C Method: Look at the Activity Series to determine if a reaction will occur Exchange A and B Write the correct product formulas with states.

Single Displacement Reactions Remember: Metals DO NOT have a CHARGE Metals are just electrically neutral atoms that share electrons Metals become CATIONS upon reacting (see charge on Activity Series)

Single Displacement Reactions Example #2 Magnesium metal is placed in a beaker full of hydrochloric acid Mg (s) + 2 HCl (aq)  MgCl2 + H2 (g) A B C Now let’s learn to assign OXIDATION NUMBERS.

Assigning Oxidation Numbers By Professor Walker Waugh

Assigning Oxidation Numbers Remember O I L R I G Oxidation Is a Loss of Electrons Reduction Gain of Electrons

Assigning Oxidation Numbers Assigning Oxidation Numbers (state) = a book keeping system to see where the electron are flowing Method Elemental forms = 0 Monatomic Ions = charge Overall Oxidation Number for a Polyatomic ions = charge Overall Oxidation Number of a compound = 0 Oxygen = -2, except in peroxide, O22-, = -1 Hydrogen in covalent compound = +1 Hydrogen in ionic compound = -1

Assigning Oxidation Numbers Example 1 Individual Oxidation Number = IOxN (written above the element) Overall Oxidation Number = OOxN (written below the element) IOxN 0 +1 -1 +2 -1 0 Mg (s) + 2 HCl (aq)  MgCl2 + H2 (g) p+ 12 p+1 p+17 p+ 12 p+ 17 p+1 e- 12 e- 0 e- 18 e- 10 e- 18 e- 1 0 +1 -1 = 0 +2 -2 = 0 0 OOxN

Assigning Oxidation Numbers Individual Oxidation Number = IOxN (written above the element) ONLY the individual oxidation numbers are reported Overall Oxidation Number = OOxN (written below the element) The overall oxidation number is the mathematical operation and should NOT be reported IOxN 0 +1 -1 +2 -1 0 Mg (s) + 2 HCl (aq)  MgCl2 + H2 (g) OOxN 0 +1 -1= 0 +2 -2= 0 0 Always to MULTIPLY by the subscript when going from the IOxN to the OOxN and Always to DIVIDE by the subscript when going from the OOxN to the IOxN.

Assigning Oxidation Numbers Example 2 Assign the oxidation number for carbon in the oxalate ion: IOxN +3 -2 C2O42_ OOxN +6 -8 = -2 Answer: The oxidation number for carbon in oxalate is +3.

Assigning Oxidation Numbers Example 2 METHOD: 1. The Overall Oxidation Number, OOxN, for a polyatomic ion is equal to the charge. IOxN C2O42_ OOxN = -2 2. Assign oxidation numbers to oxygen. That is the only other rule that you can use at this time. IOxN -2 OOxN = -2

Assigning Oxidation Numbers 3. Assign oxidation numbers to oxygen. That is the only other rule that you can use at this time. IOxN -2 C2O42_ OOxN = -2 4. To determine the OVERALL oxidation number, OOxN for an atom, you must multiple by the subscript. Thus for oxygen: -2 x 4 = -8. OOxN -8 = -2

Assigning Oxidation Numbers 5. The OVERALL oxidation number must equal -2. Therefore the OOxN for carbon is +6, since +6 – 8 = -2. IOxN -2 C2O42_ OOxN +6 -8 = -2 6. Finally, to determine the INDIVIDUAL Oxidation Number, IOxN, of carbon, DIVIDE by the OOxN of carbon by the subscript of carbon. Thus, +6 / 2 = +3. IOxN +3 -2 Answer: The oxidation number for carbon in oxalate is +3.