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1 Chapter 3: Chemical Stoichiometry Chemical Equations (Write, Balance, Interpret) Reactions You Should Know Formula Weights (Must know chemical formula Avogadro s number and the Mole Limiting Reactants Per Cent Yield Empirical Formulas Solubility Rules Oxidation Rules Chemical REACTIONS you should know COMBUSTION (Of a Hydrocarbon) C 6 H 5 OH + O 2 CO 2 + H 2 O NEUTRALIZATION (Acid + Base) HC 2 H 3 O 2 (aq) + NaOH(aq) H 2 O + NaC 2 H 3 O 2 (aq) PRECIPITATION BaCl 2 (aq) + Na 2 SO 4 (aq) BaSO 4 (s) + NaCl(aq) Chemical Reactions Combustion of Ethanol 1 st Write Reaction C 2 H 6 OH + O 2 CO 2 + H 2 O 2 nd Balance Equation 1 C 2 H 6 OH +? O 2 2 CO 2 + 7/2 H 2 O must start somewhere why not here then and What number for oxygen? A balanced chemical equation has the same number of atoms in the reactants as in the products. 4 C 2 H 6 OH + 13 O 2 8 CO H 2 O 8=C=8 28=H=28 30=O=30 3 rd Interpret Reaction The reactants must be in the ratio of 4 : 13 The balanced equation has = 39 total moles Calculate Formula Weights Must Know Formulas Use Formula Weight to Convert Grams to Moles & Moles to Grams Oxygen Ozone Methane Ammonia C 6 H 12 O 6 O 2 O 3 : 2 x 16 = 32 grams per unit : 3 x 16 = 48 grams per mole CH 4 : x 1 = 16 gms / mole NH 3 : x 1 = 17 g / mole (6 x 12) + (12 x 1) + (6 x 16) = 180 How many moles in 2.4 grams of Ozone 2.4 Grams x = moles? Grams Must know formula for Ozone! 2.4 Grams x = moles 48 Grams 1

2 Use Formula Weight to Convert Grams to Moles & Moles to Grams How many grams in 1½ moles of Methane? Grams 1.5 moles x = Grams Formula for Methane? 16 Grams 1.5 moles x = 24 Grams Use AVOGARDRO S Number to Convert Molecules to Moles & Moles to Molecules Molecules x = moles x10 Molecules Units, Units, Units 6.02x10 moles x 1 23 Molecules = Molecules moles Use Chemical Formula to Convert Molecules to Atoms & Atoms to Molecules 6.02 x 10 How many Hydrogen atoms in 1 mole of Methane? Atoms Molecul es x = Atoms 1 Molecules 23 Molecules Formula for methane? 4 x 1 hydrogen atoms = 2.41x 10 Molecules 24 hydrogen atoms Interpretation of Chemical Reactions Using Stoichiometry The chemical arithmetic necessary to relate the moles of reactants & products Chemical Reactions are Interpreted on the Mole basis but chemicals are weighed in the laboratory in Grams Yields of Chemical Reactions Chemical Reactions do not always go the way we expect them to Using stoichiometry we can calculate the theoretical (Maximum) amount of product formed in a reaction. If 10.0 moles of NO are reacted with 5.0 mole O 2, how many moles NO 2 are produced? 2 NO (g) + O 2 (g) 2 NO 2 (g) (a) 2.0 mol NO 2 (b) 6.0 mol NO 2 (c) 10.0 mol NO 2 (c) 10.0 mol NO (d) 16.0 mol NO 2 2 (e) 32.0 mol NO 2 2

3 The Limiting Reactant A reaction stops when one reactant is totally consumed. This is the limiting reactant. The other reactants are excess reactants. If 10.0 moles of NO are reacted with 6.0 moles O 2, how many moles NO 2 are produced? 2 NO (g) + O 2 (g) 2 NO 2 (g) (a)what LIMITS the reaction? (b)what is in excess? The amount of NO The amount of Oxygen If 10.0 moles of NO are reacted with 6.0 moles O 2, how many moles of the excess 1.0 mol O mol O mol NO 8.0 mol NO reagent remain? 2 NO (g) + O 2 (g) 2 NO 2 (g) 1.0 mol O 2 Yields of Chemical Reactions Using stoichiometry the theoretical (Maximum) amount of product formed is calculated. If the actual amount of product formed is less than the theoretical amount, the percentage yield is calculated. Actual product yield % yield = 100% Theoretical product yield With a 50 % Yield, How many moles of NH 3 are produced from 1 ½ moles of H 2 and 1 mole of N 2? 1 st Write and Balance Reaction 3 H N 2 2 NH 3 or 1 ½ H 2 + ½ N 2 1 NH 3 So ½ mole NH 3 formed with ½ mole of Nitrogen in excess If grams Ozone reacts with grams of nitrogen monoxide, to produce nitrogen dioxide 1. How many grams of product will be formed? 2. What is the limiting reagent? 3. How much excess reagent will remain? 1 st Write & Balance Reaction : 1 O NO 3 NO 2 6 = O = 6 3 = N = 3 3

4 0.720 grams O 3 & grams of NO For O : grams = moles 48 grams 3 O 3 1 mole For NO have grams = moles 30.0 grams 1 O NO 3 NO 2 1 (0.015) O (0.015) NO 3 (0.015) NO O NO NO O NO NO 2 Have mol NO Need mole NO Therefore Nitrogen monoxide is limiting reagent 1 moles O3 needed = moles = grams grams O3 needed = moles = mole Grams O 3 excess = = Two Approaches to Chemical Formulas 1. Given Formula Determine % Composition of Each Element in Compound. What is the % H & O in hydrogen peroxide? 2. Determine Formula Given % Composition of Each Element in Compound What is the Formula of the compound that is 5.88% H & %O? Find Empirical Formula Weight K = Weight O = Total Weight = First determine % by Weight % by Wt K = x 100 = % by wt K and 16.6 % by wt O Choose any total weight 100 grams is convenient Construct the following work sheet Element % Wt moles ratio K g 83.4 g / 39 = 2.14 = 2 O g 16.6 g / 16 = 1.04 = 1 Therefore formula is K 2 O Chapter 4 Reactions in Aqueous Solutions & Solution Stoichiometry 4

5 WHAT IS A SOLUTION? A HOMOGENEOUS MIXTURE OF TWO OR MORE SUBSTANCES Solvent + Solute = Solution Concentrations in Solutions Molarity ELECTROLYTE A substance that dissolves in water to produce IONS Example: HCl(aq), NaOH(aq), NaCl(aq) NON ELECTROLYTE A substance that DOES NOT produce IONS, remain as molecules, when dissolved in water. Example: sugar(aq), Ethylene glycol (aq) What do the following compounds do when mixed with water? {are they water soluble?) Magnesium Iodide Aluminum Nitrate Ammonium Sulfate Perchloric Acid Silver Chloride MgI 2 Mg 2+ (aq) + 2 I - (aq) Al(NO 3 ) 3 Al 3+ (aq) + 3NO - 3 (aq) (NH 4 ) 2 SO 4 2NH + 4 (aq) + SO4 2- (aq) HClO 4 H + (aq) + ClO - 4 (aq) AgCl Is Insoluble WHY DOES A REACTION OCCUR? 1. Precipitation Reaction 2. Neutralization Reaction 3. Oxidation-Reduction Reaction An INSOLUBLE Solid is Formed Acid-Base slightly ionized Substance water Formed Change in Oxidation State Will precipitation occur when the following solutions are mixed? If so, write net ionic equation for reaction (a) Na 2 CO 3 and Ag C 2 H 3 O 2.. Yes 2 Ag + (aq) + CO 2-3 (aq) Ag 2 CO 3 (s) (b) NaNO 3 and NiSO 4. No reaction (c) FeSO 4 and Pb(ClO 4 ) 2.Yes Pb 2+ (aq) + SO 2-4 (aq) PbSO 4 (s) 5

6 CHEMICAL EQUATIONS MOLECULAR - Uses The Full Formula Of Reactants and Products IONIC ONLY Ions Are Shown NET IONIC SPECTATOR Ions Are Removed From Ionic Equation SPECTATOR Ions Ions that do not take part in the chemical reaction Equations for Silver Perchlorate + Sodium Nitrate The Molecular Equation is : AgClO 4 (aq) + NaNO 3 (aq) AgNO 3 (aq) + NaClO 4 (aq) The Ionic Equation is : Ag + (aq) + ClO 4 - (aq) + Na + (aq) + NO 3 - (aq) Ag + (aq) + NO 3 - (aq) + Na + (aq) + ClO 4 - (aq) Spectator Ions: all are spectator ions There is no Net Ionic Equation because there is NO REACTION!!!!!! Write (1) Molecular (2) Ionic & (3) Net Ionic Equations for H 2 SO 4 (aq) + KOH(aq)????? H 2 O(liq) + K 2 SO 4 (aq) 2 H + (aq) + SO 2-4 (aq) + K + (aq) + OH - (aq) H 2 O(liq) + K + (aq) + SO 2-4 (aq) 2 H + (aq) + OH - (aq) H 2 O(liq) Equations for Sulfuric Acid + Potassium Hydroxide The Molecular Equation is : H 2 SO 4 (aq) + 2 KOH(aq) H 2 O(liq) + K 2 SO 4 (aq) The Ionic Equation is : 2 H + (aq) + SO 4 2- (aq) + 2 K + (aq) + OH - (aq) Spectator Ions K + & SO 4-2 H 2 O(liq) + 2 K + (aq) + SO 4 2- (aq) Net Ionic Equation is : 2 H + (aq) + OH - (aq) H 2 O (liq) Write (1) Molecular (2) Ionic & (3) Net Ionic Equations for Hg(NO 3 ) 2 (aq) + NaI(aq)??????? NaNO 3 (aq) + HgI 2 (S) Hg +2 (aq) + NO - 3 (aq) + Na + (aq) + I - (aq) Na + (aq) + NO - 3 (aq) + HgI 2 (S) Hg +2 (aq) + 2 I - ( aq) HgI 2 (S) Equations for Mercury II Nitrate + Sodium Iodide The Molecular Equation is : Hg(NO 3 ) 2 (aq) + NaI (aq) NaNO 3 (aq) + HgI 2 (s) The Ionic Equation is : Hg +2 (aq) + NO 3 - (aq) + Na + (aq) + I - (aq) Spectator Ions Na + & NO 3 - Na + (aq) + NO 3 - (aq) + HgI 2 (s) Net Ionic Equation is : Hg +2 (aq) + 2 I - (aq) HgI 2 (s) 6

7 How do you recognize an Oxidation Reduction Reaction? H 2 (g) + O 2 (g) H 2 O (g) On one side of element all by its self While on the other side element in combination Another example Zn (s) + 2 HCl (aq) H 2 (g) + ZnCl 2 (aq) and Assigning Oxidation Numbers Must Know Oxidation RULES See Text / Lab Manual 1. EACH ATOM in a PURE ELEMENT has an Oxidation Number of ZERO H 2, O 2, N 2, F 2, Na, Fe, Zn etc For example Sodium...Na Atomic Number.11 Number of Protons 11 Number of Electrons..11 Same number of protons and electrons 2. Alkali Metals always = +1 {Li, Na, K, Rb, Cs 3. Alkaline Earth Metals always = +2 {Mg, Ca, Sr, Ba 4. The oxidation number of H in its compounds = + 1 Most of the time: H 2 O ; Na OH ; HCl ; H C 2 H 3 O 2 But Not all the time Ca H 2 {H = -1 in Hydrides 5. Oxidation number of F in compounds always = Oxidation number of O in its compounds = 2 Most of the time: H 2 O ; Na OH ; H C 2 H 3 O 2 But Not all the time H 2 O 2 {O = -1 in Peroxides OXIDATION REDUCTION What is The OXIDATION NUMBER For EACH Element Present? What is OXIDIZED? What is REDUCED? What is The OXIDIZING AGENT? What is The REDUCING AGENT? Types of Equations (a) Molecular (b) Ionic & (c) Net Ionic (a) Molecular: the formula for each substance are written in the molecular form Zn (s) + 2 HBr (aq) H 2 (g)+ ZnBr 2 (aq) (b) Ionic: shows dissolved species in ionic form Zn(s) + 2H + (aq) + 2Br - (aq) H 2 (g) + Zn 2+ (aq) + 2Br - (aq) (c) Net Ionic: only shows the ionic species that actually take part in the reaction Zn (s) + 2 H + (aq) H 2 (g) + Zn 2+ (aq) Redox Reactions Zn(s) + HCl(aq) H 2 (g) + ZnCl 2 (aq) Oxidation Numbers What is OXIDIZED What is REDUCED Oxidizing AGENT? Reducing AGENT? Zn(s) & H 2 (g) = 0 Zn in ZnCl 2 (aq) = +2 Cl in HCl and ZnCl 2 = -1 Zn 0 Zn 2+ H + H 2 O HCl(aq) Zn metal 7

8 Putting Concepts Together A sample of 70.5 mg of potassium phosphate is added to 15.0 ml of M silver nitrate, (a) Write the molecular equation for the reaction. (b) Write ionic and net ionic equations (c) What is the limiting reactant in the reaction? (d) Calculate the theoretical yield, in grams, of the precipitate that forms. Potassium phosphate s chemical formula is K 3 PO 4 Silver nitrate s chemical formula is AgNO 3 Molecular Equation K 3 PO 4 (aq) + AgNO 3 (aq) Ag 3 PO 4 (?) + KNO 3 (?) According to solubility guidelines Ag 3 PO 4 will precipitate while K 3 PO 4, AgNO 3, & KNO 3 are soluble Complete ionic equation K + (aq) + PO 4 3 (aq) + Ag + (aq), + NO 3 (aq) Ag 3 PO 4 (s) + K + (aq) + NO 3-1 (aq) ionic equation K + (aq) + PO 3 4 (aq) + Ag + (aq) + NO 3 (aq) Ag 3 PO 4 (s) + K + (aq) + NO -1 3 (aq) (b) Net Ionic Equation Ag + (aq) + PO 4 3 (aq) Ag 3 PO 4 (s) Spectator ions: K + & NO 3 ions (c) To determine the limiting reactant, must determine the number of moles of each reactant. According to the balanced equation, 1 K 3 PO 4 (aq) + 3AgNO 3 (aq) Ag 3 PO KNO 3 (aq) 1 mol K 3 PO 4 requires 3 mol of AgNO 3. There are moles of K 3 PO 4 and moles of AgNO 3 Rebalance equation 3.32 K 3 PO AgNO Ag 3 PO KNO 3 Not enough AgNO 3 (have 7.5x 10-4 moles) for the K 3 PO 4 therefore AgNO 3 is the limiting reactant. Molarity (M) Memorize the definition moles of solute Molarity = Liters of solution or Moles Solute = Molarity x Liters of solution moles = M x V How are Moles determined from Molarity? Moles of Solute = Molarity x (Volume in Liters) Calculate the number of moles of HCl in 50.0 ml of 2.00 M HCl(aq) Moles = M x V = (0.0500)x(2.00) = Calculate the number of moles of NaOH in 51.0 ml of 2.00 M NaOH Moles = M x V = (0.0510)x(2.00) =

9 Write & Balance Reaction 1 HCl (aq) + 1 NaOH (aq) 1 H 2 O + 1 NaCl (aq) or 0.1HCl(aq) NaOH(aq) 0.1H 2 O + 0.1NaCl(aq) What is the Limiting Reagent? What is the Theoretical Yield of water? grams of NaCl formed, % Yield is HCl(aq) mole = 1.8 grams g NaCl = 0.1 mole Therefore 100% yield Preparing solutions By Dilution Start with 50.0 ml of 2.00 M HCl(aq) Add water until have ml of solution What is the Molarity of the new solution? (1) M 1 V 1 = M 2 V 2 = moles of Solute (2) (2.00)(50.0) = ( M 2 )(100.0) =??? (3) M 2 = 1.00 Chap 5 THERMO chemistry Review HEAT CHANGES that take place during 1. Physical Changes (Change of Phase) & 2. Chemical Changes (Chemical Reactions) Energy cannot be created or destroyed BUT it can be transformed from one form to another (potential & kinetic) or transferred between system and surroundings One of the basic assumptions of thermodynamics is the idea that the Universe can be divided into a System and its Surroundings The study of energy and its transformation is known as THERMODYNAMICS THERMOCHEMISTRY is the portion of thermodynamics that pertains to chemical reactions The energy possessed by a system is called its internal energy internal energy of a system is the sum of all kinetic and potential energies of all components of the system Note: Most books use U for internal energy. This book uses the older nomenclature E The 1 st Law of Thermodynamics states that the energy of the universe is constant U universe = U system - U surrounding = 0 Energy can be transferred from system to surroundings (or vice versa) but it can't be created or destroyed U system = U surrounding A more useful form of the first law: U system = q + w The change in the internal energy of a system is equal to the sum of the heat.(gained or lost by the system) and the work...(done by or on the system) U system = q + w For q For w For U sys + means system GAINS heat (endothermic) - means system LOSES heat (exothermic) means work done ON system - means work done BY system means system GAINS energy - means system LOSES energy 9

10 Changes involving energy can either be PHYSICAL or CHEMICAL PHYSICAL Changes 1. Change in state (No temperature change) 2. Change within a state (Change in temp) CHEMICAL Changes (Reactions 1. Formation 2. Combustion 3. Neutralization 10

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