1 IB Topic 1: Quantitative Chemistry 1.3: Chemical Equations  1.3.1 Deduce chemical equations when all reactants and products are given  1.3.2 Identify.

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1 IB Topic 1: Quantitative Chemistry 1.3: Chemical Equations  Deduce chemical equations when all reactants and products are given  Identify the mole ratio of any two species in a chemical reaction.  Apply the state symbols (s), (l), (g) and (aq)

Deduce chemical equations when all reactants and products are given. What is a chemical reaction? Rearrangement of atoms forming new substances Reactants  Products Some reactions are desirable… Glucose + oxygen  Carbon dioxide + water …some are not. Iron + oxygen  iron (III) oxide (a.k.a. rust)

3 Chemical Reactions  A few ways to determine whether a chemical reaction has taken place: Heat is absorbed or given off Change in color Change in odor Production of a gas or solid (precipitate) Not easily reversible (it won’t recreate the reactants)

4

Deduce chemical equations when all reactants and products are given. Chemical formulas are easier and more informative to use in equations than words. A skeleton equation is a chemical equation that shows what reactants and products are involved. It does not necessarily indicate the relative amounts of reactants and products. Fe + O 2  Fe 2 O 3 Sometimes the skeleton equation is balanced (all coefficients = 1). SnO 2 (s)  Sn(s) + O 2 (g)

6 Identify types of chemical reactions Types of Chemical Reactions 1) Synthesis or Combination Reaction: Two or more substances combine to form a single substance.  2Na(s) + Cl 2 (g)  2NaCl(s)  CaO(s) + H 2 O(l)  Ca(OH) 2 (aq)  2H 2 (g) + O 2 (g)  2H 2 O (l)  4Fe(s) + 3O 2 (g)  2Fe 2 O 3 (s)  N 2 (g) + 3H 2 (g)  2NH 3 (g)

7 Identify types of chemical reactions Types of Chemical Reactions 2)Decomposition Reactions: A single substance is broken down into two or more substances.  CaCO 3 (s)  CaO(s) + CO 2 (g)  2H 2 O(l)  2H 2 (g) + O 2 (g)  2H 2 O 2 (aq)  2H 2 O(l) + O 2 (g)

8 Identify types of chemical reactions Types of Chemical Reactions 3) Single-Replacement Reactions: One element replaces another in a compound.  Mg(s) + Zn(NO 3 ) 2 (aq)  Mg(NO 3 ) 2 (aq) + Zn(s)  2K(s) + 2H 2 O(l)  2KOH (aq) + H 2 (g)  Cu(s) + 2AgNO 3 (aq)  Cu(NO 3 ) 2 (aq) + 2Ag(s)  2Al(s) + Fe 2 O 3 (s)  Al 2 O 3 (s) + 2Fe(s)  Cl 2 (g) + 2NaBr(aq)  2NaCl(aq) + Br 2 (g)

9 Identify types of chemical reactions Types of Chemical Reactions 4) Double-Replacement Reactions: Ions of two reacting compounds trade places.  Na 2 CO 3 (aq) + CaCl 2 (aq)  CaCO 3 (s) + 2NaCl(aq)  Na 2 S(aq) + Cd(NO 3 ) 2 (aq)  CdS(s) + 2NaNO 3 (aq)  NaOH(aq) + HCl(aq)  NaCl(aq) + H 2 O(l)  H 2 SO 4 (aq) + 2NaCN(aq)  Na 2 SO 4 (aq) + 2HCN(g)

10 Identify types of chemical reactions Types of Chemical Reactions 5) Combustion Reactions: A compound reacts with oxygen. Products are usually carbon dioxide and water.  CH 4 (g) + O 2 (g)  CO 2 (g) + H 2 O(g)  C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(g)  2C 8 H 18 (l) + 25O 2 (g)  16CO 2 (g) + 18H 2 O(g)  C 2 H 5 OH(l) + 3O 2 (g)  2CO 2 (g) + 3H 2 O(g)

Deduce chemical equations when all reactants and products are given. Balancing chemical equations The Law of Conservation of Matter: In a chemical (non-nuclear) reaction, atoms are neither created nor destroyed. For an equation to be balanced the number of atoms of each element is the same on both sides of the equation. H 2 (g) + O 2 (g)  H 2 O (l)(unbalanced) 2H 2 (g) + O 2 (g)  2H 2 O (l)(balanced) K(s) + H 2 0(l)  KOH (aq) + H 2 (g)(unbalanced) 2K(s) + 2H 2 0(l)  2KOH (aq) + H 2 (g)(balanced) C 6 H 12 O 6 + O 2  CO 2 + H 2 O(unbalanced) C 6 H 12 O 6 + 6O 2  6CO 2 + 6H 2 O(balanced)

12 Counting Atoms  Subscripts indicate how many of a specific atom is present in a compound This is the small number to the bottom right of an element in a compound. Ex. H 2 O… 2 is the subscript  Coefficients tell us how many units of the compound we have Ex. 3 H 2 O means that we have 3 water molecules So, how many hydrogen atoms do we have? Oxygen atoms?  Let’s practice

Deduce chemical equations when all reactants and products are given. Rules for Balancing Equations 1)Write the correct formulas for the reactants on the left side of the yield sign and products on the right side. 2)Count the number of atoms of each element in the products and the reactants. 3)Balance the elements one at a time by using coefficients. Do not change the subscripts in the chemical formulas. 4)Check each atom or polyatomic ion to make sure the equation is balanced. 5)Make sure all coefficients are in the lowest possible ratio.

Deduce chemical equations when all reactants and products are given. Rules for Balancing Equations 1)Write the correct formulas for the reactants on the left side of the yield sign and products on the right side. 2)Count the number of atoms of each element in the products and the reactants. 3)Balance the elements one at a time by using coefficients. Do not change the subscripts in the chemical formulas. 4)Check each atom or polyatomic ion to make sure the equation is balanced. 5)Make sure all coefficients are in the lowest possible ratio. N 2 + H 2  NH 3 1) Formulas given 2) ReactantsProducts 2N 2H  1N 3H 3) Balance N by putting 2 in front of NH 3 N 2 + H 2  2NH 3 Balance H by putting 3 in front of H 2 N 2 + 3H 2  2NH 3 4) 2N 6H  2N 6H 5) Lowest ratio of coefficients

Deduce chemical equations when all reactants and products are given. Rules for Balancing Equations 1)Write the correct formulas for the reactants on the left side of the yield sign and products on the right side. 2)Count the number of atoms of each element in the products and the reactants. 3)Balance the elements one at a time by using coefficients. Do not change the subscripts in the chemical formulas. 4)Check each atom or polyatomic ion to make sure the equation is balanced. 5)Make sure all coefficients are in the lowest possible ratio. KClO 3  KCl + O 2 1) Formulas given 2) Reactants Products 1K 1Cl 3O  1K 1Cl 2O 3) Balance O by putting 2 in front of KClO 3 and 3 in front of O 2 2KClO 3  KCl + 3O 2 Balance K & Cl by putting 2 in front of KCl 2KClO 3  2KCl + 3O 2 4) 2K 2Cl 6O  2K 2Cl 6O 5) Lowest ratio of coefficients

Deduce chemical equations when all reactants and products are given. Balance the following equations  Fe(s) + O 2 (g)  Fe 2 O 3 (s)  CaCO 3 (s)  CaO(s) + CO 2 (g)  Al(s) + Fe 2 O 3 (s)  Al 2 O 3 (s) + Fe(s)  H 2 SO 4 (aq) + NaCN(aq)  Na 2 SO 4 (aq) + HCN(g)  C 2 H 5 OH(l) + O 2 (g)  CO 2 (g) + H 2 O(g)

Deduce chemical equations when all reactants and products are given. Balance the following equations  4Fe(s) + 3O 2 (g)  2Fe 2 O 3 (s)  CaCO 3 (s)  CaO(s) + CO 2 (g)  2Al(s) + Fe 2 O 3 (s)  Al 2 O 3 (s) + 2Fe(s)  H 2 SO 4 (aq) + 2NaCN(aq)  Na 2 SO 4 (aq) + 2HCN(g)  C 2 H 5 OH(l) + 3O 2 (g)  2CO 2 (g) + 3H 2 O(g)

Deduce chemical equations when all reactants and products are given. Aluminum bromide + chlorine yield aluminum chloride + bromine 2AlBr 3 + 3Cl 2  2AlCl 3 + 3Br 2 Seven elements are diatomic: H 2, N 2, O 2, F 2, Cl 2, Br 2, I 2 Copper + oxygen produces copper(I) oxide 4Cu + O 2  2Cu 2 O

Deduce chemical equations when all reactants and products are given. Sodium chlorate decomposes to sodium chloride and oxygen gas Aluminum nitrate plus sodium hydroxide yields aluminum hydroxide plus sodium nitrate Ethane (C 2 H 6 ) burns in oxygen to produce carbon dioxide and water vapor

Deduce chemical equations when all reactants and products are given. Sodium chlorate decomposes to sodium chloride and oxygen gas 2NaClO 3  2NaCl + 3O 2 Aluminum nitrate plus sodium hydroxide yields aluminum hydroxide plus sodium nitrate Al(NO 3 ) 3 + 3NaOH  Al(OH) 3 + 3NaNO 3 Ethane (C 2 H 6 ) burns in oxygen to produce carbon dioxide and water vapor 2C 2 H 6 + 7O 2  4CO 2 + 6H 2 O

Apply the state symbols (s), (l), (g) and (aq) Symbols used in chemical reactions + Used to separate two reactants or products A “Yields” arrow (  ) separates products from reactants  Used in place of  for reversible reactions (s)Designates a solid reactant or product (l)Designates a liquid reactant or product (g)Designates a gaseous reactant or product (aq)Designates an aqueous reactant or product  Indicates that heat is supplied to the reaction

Apply the state symbols (s), (l), (g) and (aq) (s)Designates a solid reactant or product (l)Designates a liquid reactant or product (g)Designates a gaseous reactant or product (aq)Designates an aqueous reactant or product

Apply the state symbols (s), (l), (g) and (aq)  Rule 1. All compounds of Group IA elements (the alkali metals) are soluble.  For example, NaNO 3, KCl, and LiOH are all soluble compounds. This means that an aqueous solution of KCl really contains the predominant species K + and Cl - and, because KCl is soluble, no KCl is present as a solid compound in aqueous solution: KCl (s) => K + (aq.) + Cl - (aq.)

Apply the state symbols (s), (l), (g) and (aq)  Rule 2. All ammonium salts (salts of NH 4 + ) are soluble.  For example, NH 4 OH is a soluble compound. Molecules of NH 4 OH completely dissociate to give ions of NH 4 + and OH - in aqueous solution.

Apply the state symbols (s), (l), (g) and (aq)  Rule 3. All nitrate (NO 3 - ), chlorate (ClO 3 - ), perchlorate (ClO 4 - ), and acetate (CH 3 COO - or C 2 H 3 O 2 - are soluble.  For example, KNO 3 would be classified as completely soluble by rules 1 and 3. Thus, KNO 3 could be expected to dissociate completely in aqueous solution into K + and NO 3 - ions: KNO 3 => K + (aq. ) + NO 3- (aq.)

Apply the state symbols (s), (l), (g) and (aq)  Rule 4. All chloride (Cl - ), bromide (Br - ), and iodide (I - ) salts are soluble except for those of Ag +, Pb 2+, and Hg  For example, AgCl is a classic insoluble chloride salt: AgCl (s) Ag + (aq.) + Cl - (aq.) (K sp = 1.8 x ).

Apply the state symbols (s), (l), (g) and (aq)  Rule 5. All sulfate ( SO 4 = ) compounds are soluble except those of Ba 2+, Sr 2+, Ca 2+, Pb 2+, Hg 2 2+, and Hg 2+. Ca 2+ and Ag + sulfates are only moderately soluble.  For example, BaSO 4 is insoluble (only soluble to a very small extent): BaSO 4(s) Ba 2+ (aq.) + SO 4 2- (aq.) (K sp = 1.1 x ). Na 2 SO 4 is completely soluble: Na 2 SO 4(s) => 2 Na + (aq.) + SO 4 2- (aq.).

Apply the state symbols (s), (l), (g) and (aq)  Rule 6. All hydroxide (OH - ) compounds are insoluble except those of Group I-A (alkali metals) and Ba 2+, Ca 2+, and Sr 2+.  For example, Mg(OH) 2 is insoluble (Ksp = 7.1 x 10-12) NaOH and Ba(OH) 2 are soluble, completely dissociating in aqueous solution: NaOH(s) => Na + (aq.) + OH - (aq.), a strong base Ba(OH) 2 (s) => Ba 2+ (aq.) + 2OH - (aq.) (Ksp = 3 x 10-4)

Apply the state symbols (s), (l), (g) and (aq)  Rule 7. All sulfide (S 2- ) compounds are insoluble except those of Groups I-A and II-A (alkali metals and alkali earths).  For example, Na 2 S(s) 2Na + (aq.) + S 2- (aq.) MnS is insoluble (Ksp = 3 x 10-11).

Apply the state symbols (s), (l), (g) and (aq)  Rule 8. All sulfites (SO 3 2- ), carbonates (CO 3 2- ), chromates (CrO 4 2- ), and phosphates (PO 4 3- ) are insoluble except for those of NH 4 + and Group I-A (alkali metals)(see rules 1 and 2).  For example, calcite, CaCO 3 (s) Ca 2+ (aq.) + CO 3 2- (aq.) (Ksp = 4.5 x 10-9).

Apply the state symbols (s), (l), (g) and (aq)  OR... Use a Solubility Chart  Let’s use the following reactions to apply this assessment statement 1. Hydrogen peroxide decomposes into water and oxygen gas 2. Potassium phosphate reacts with magnesium chloride to form magnesium phosphate and potassium chloride 3. Iron reacts with copper (II) sulfate to form iron (II) sulfate and copper 4. Sulfuric acid (hydrogen sulfate) reacts with copper (II) oxide to produce copper (II) sulfate and water

32

Apply the state symbols (s), (l), (g) and (aq) 1. Hydrogen peroxide decomposes into water and oxygen gas 2 H 2 O 2 (aq)  2 H 2 O(l) + O 2 (g) 2. Potassium phosphate reacts with magnesium chloride to form magnesium phosphate and potassium chloride 2K 3 PO 4 (aq) + 3MgCl 2 (aq)  Mg 3 (PO 4 ) 2 (s) + 6KCl (aq) 3. Iron reacts with copper (II) sulfate to form iron (II) sulfate and copper Fe(s) + CuSO 4 (aq)  FeSO 4 (aq) + Cu(s) 4. Sulfuric acid (hydrogen sulfate) reacts with copper (II) oxide to produce copper (II) sulfate and water H 2 SO 4 (aq) + CuO (s)  CuSO 4 (aq) + H 2 O (l)

34 Cookies and Chemistry…Huh!?!?  Just like chocolate chip cookies have recipes, chemists have recipes as well  Instead of calling them recipes, we call them reaction equations  Furthermore, instead of using cups and teaspoons, we use moles  Lastly, instead of eggs, butter, sugar, etc. we use chemical compounds as ingredients

Identify the mole ratio of any two species in a chemical reaction. Coefficients are in particles or in moles 2 f.u. NaClO 3  2 f.u. NaCl + 3 molecules O 2 2 mol NaClO 3  2 mol NaCl + 3 mol O 2 We will use moles because we measure moles in grams The coefficients give us mole ratios Mole ratio NaClO 3 :NaCl is 2:2 Mole ratio NaCl:O 2 is 2:3 Mole ratio O 2 :NaCl is 3:2

Identify the mole ratio of any two species in a chemical reaction. Consider the following equation 2K 2 Cr 2 O 7 + 2H 2 O + 3S  4KOH + 2Cr 2 O 3 + 3SO 2 What is the KOH:S mole ratio? What is the K 2 Cr 2 O 7 :Cr 2 O 3 mole ratio? What is the mole ratio between sulfur dioxide and water? What species will give mole ratios of 4:2?

Identify the mole ratio of any two species in a chemical reaction. Consider the following equation 2K 2 Cr 2 O 7 + 2H 2 O + 3S  4KOH + 2Cr 2 O 3 + 3SO 2 What is the KOH:S mole ratio? 4:3 What is the K 2 Cr 2 O 7 :Cr 2 O 3 2:2 What is the mole ratio between sulfur dioxide and water? 3:2 What species will give mole ratios of 4:2? KOH:K 2 Cr 2 O 7 KOH:H 2 OKOH:Cr 2 O 3

Identify the mole ratio of any two species in a chemical reaction.  Looking at a reaction tells us how much of something you need to react with something else to get a product (like the cookie recipe)  Be sure you have a balanced reaction before you start!  Example: 2 Na + Cl 2  2 NaCl  This reaction tells us that by mixing 2 moles of sodium with 1 mole of chlorine we will get 2 moles of sodium chloride  What if we wanted 4 moles of NaCl? 10 moles? 50 moles?

Identify the mole ratio of any two species in a chemical reaction.  These mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemical  Example: How many moles of chlorine is needed to react with 5 moles of sodium (without any sodium left over)? 2 Na + Cl 2  2 NaCl 5 moles Na 1 mol Cl 2 2 mol Na = 2.5 moles Cl 2

Identify the mole ratio of any two species in a chemical reaction.  Looking at a reaction tells us how much of something you need to react with something else to get a product (like the cookie recipe)  Be sure you have a balanced reaction before you start!  Example: 2 Na + Cl 2  2 NaCl  This reaction tells us that by mixing 2 moles of sodium with 1 mole of chlorine we will get 2 moles of sodium chloride  What if we wanted 4 moles of NaCl? 10 moles? 50 moles?

 How many moles of sodium chloride will be produced if you react 2.6 moles of chlorine gas with an excess (more than you need) of sodium metal? 2 Na + Cl 2  2 NaCl 2.6 moles Cl 2 2 mol NaCl 1 mol Cl 2 = 5.2 moles NaCl Identify the mole ratio of any two species in a chemical reaction.

Practice  Write the balanced reaction for hydrogen gas reacting with oxygen gas. 2 H 2 + O 2  2 H 2 O How many moles of reactants are needed? What if we wanted 4 moles of water? What if we had 3 moles of oxygen, how much hydrogen would we need to react and how much water would we get? What if we had 50 moles of hydrogen, how much oxygen would we need and how much water produced? 2 mol H 2 1 mol O 2 4 mol H 2 2 mol O 2 6 mol H 2, 6 mol H 2 O 25 mol O 2, 50 mol H 2 O