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UNIT 6. Overview Reactions Write formula/word equations Balance Equations Identify Types Predict Products Write Net Ionic Equations Stoichiometry Conversions.

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Presentation on theme: "UNIT 6. Overview Reactions Write formula/word equations Balance Equations Identify Types Predict Products Write Net Ionic Equations Stoichiometry Conversions."— Presentation transcript:

1 UNIT 6

2 Overview Reactions Write formula/word equations Balance Equations Identify Types Predict Products Write Net Ionic Equations Stoichiometry Conversions Limiting & Excess Reagent Percent Yield

3 Chemical Reactions Process in which one or more pure substances are converted into one or more different pure substances Process in which one or more pure substances are converted into one or more different pure substances Reactants: Zn + I 2 Product: Zn I 2

4 Indications of a Reaction Temperature Change Color Change Production of gas Formation of a precipitate Production of light

5 Chemical Equations 4 Al(s) + 3 O 2 (g) ---> 2 Al 2 O 3 (s) Reactants products Reactants react to produce products physical states The letters (s), (g), (l), and (aq) are the physical states of compounds. aq represents aqueous meaning dissolved in water (solution) coefficients. The numbers in the front are called coefficients. Subscripts represent the number of each atom in a compound Subscripts represent the number of each atom in a compound (Reactants)(Products)

6 Chemical Reactions Symbol Meaning + used to separate one reactant or product from another used to separate the reactants from the products - it is pronounced "yields" or "produces" when the equation is read used when the reaction can proceed in both directions - this is called an equilibrium arrow and will be used later in the course an alternative way of representing a substance in a gaseous state an alternative way of representing a substance in a solid state indicates that heat is applied to make the reaction proceed

7 Diatomic Elements Elements that cannot exist by themselves (always occur in pairs) Bromine (Br 2 ) Iodine (I 2 ) Nitrogen (N 2 ) Chlorine (Cl 2 ) Hydrogen (H 2 ) Oxygen (O 2 ) Fluorine (F 2 )

8 Writing Equations Practice 1.When lithium hydroxide pellets are added to a solution of sulfuric acid, lithium sulfate and water are formed. 2 LiOH (s) + H 2 SO 4(aq) Li 2 SO 4(aq) + 2 H 2 O (l) 2.When crystalline C 6 H 12 O 6 is burned in oxygen, carbon dioxide and water vapor are formed. C 6 H 12 O 6(s) + 6 O 2(g) 6 CO 2(g) + 6 H 2 O (g)

9 Balancing Equations Law of Conservation of Mass Law of Conservation of Mass Matter cannot be destroyed (atoms of reactants must equal products) Balance equations to get same number of each atom on the left and right in an equation 2 Hg atoms, 2 O atoms 2HgO(s) ---> 2 Hg(l) + O 2 (g)

10 Balancing Equations ___ Al(s) + ___ Br 2 (l) ---> ___ Al 2 Br 6 (s) __C 3 H 8 (g) + __ O 2 (g) __ CO 2 (g) + __ H 2 O(g) __ B 4 H 10 (g) + __ O 2 (g) __ B 2 O 3 (g) + __ H 2 O(g)

11 6 Types of Reactions Synthesis (combination) Decomposition Single Replacement (displacement) Double Replacement (precipitation) Combustion Acid-Base Neutralization

12 Synthesis (Combination) Reactions Two or more substances combine to form a new compound. A + X AX Synthesis of: Binary compounds H 2 + O 2 H 2 O Metal carbonates CaO + CO 2 CaCO 3 Metal hydroxides CaO + H 2 O Ca(OH) 2 Metal chlorates KCl + O 2 KClO 3 Oxyacids CO 2 + H 2 O H 2 CO 3

13 Decomposition Reactions A single compound breaks down into two or more simpler substances AX A + X Decomposition of: Binary compounds H 2 O H 2 + O 2 Metal carbonates CaCO 3 CaO + CO 2 Metal hydroxides Ca(OH) 2 CaO + H 2 O Metal chlorates KClO 3 KCl + O 2 Oxyacids H 2 CO 3 CO 2 + H 2 O

14 Single Replacement (displacement) Reactions One element replaces another in a reaction Metals replace metals Nonmetals replace nonmetals A + BX AX + B BX + Y BY + X

15 Activity Series Decide whether or not one element will replace another Metals can replace other metals provided that they are above the metal that they are trying to replace If the metal is not above what it is trying to replace, the result is no reaction

16 Double Replacement (Precipitation) Reactions Two elements or ions switch partners AX + BY AY + BX precipitateinsoluble gas molecular compound One of the compounds formed is usually a precipitate, an insoluble gas that bubbles out of solution, or a molecular compound, usually water.

17 Solubility Solubility Solubility – ability to dissolve In a double replacement (precipitate) reaction, one of the products must be insoluble in water and form a precipitate Precipitate Precipitate – insoluble solid formed by a reaction in solution If both products are soluble the result is no reaction Solubility rules help you determine whether or not a compound will form a precipitate or remain an aqueous solution

18 Solubility Rules Soluble Ionic CompoundsExcept with: Alkali metals, NH 4 + NO 3 -, C 2 H 3 O 2 -, ClO 3 -, ClO 4 - (no exceptions) Cl -, Br -, I - Ag +, Hg 2 +2, Pb +2 SO 4 -2 Sr +2, Ba +2, Ca +2, Ag +, Pb +2, Hg 2 +2 Insoluble Ionic CompoundsExcept with: CO 3 -2, PO 4 -3, SiO 3 -2, O -2, SO 3 -2, CrO 4 -2 NH 4 +, alkali metals S -2 NH 4 +, alkali metals Ca +2, Sr +2, Ba +2, Mg +2 (group 2) OH - NH 4 +, alkali metals, (Ca +2, Ba +2, Sr +2 are slightly soluble)

19 Combustion Reactions A substance combines with oxygen, releasing a large amount of energy in the form of light and heat. Produces a flame Fuel + oxygen produces carbon dioxide and water vapor C x H x + O 2 CO 2 + H 2 O

20 Acid-Base Neutralization Reactions When the solution of an acid and solution of a base are mixed Products have no characteristics of either the acid or the base Acid + Base (metal hydroxide) salt + water Salt comes from cation of base and anion of acid HY + XOH XY + H 2 O

21 Chemical Equations Molecular Equation Molecular Equation – shows complete chemical formulas of reactants and products Pb(NO 3 ) 2 (aq) + 2KI(aq) PbI 2 (s) + 2KNO 3 (aq) Complete Ionic Equation Complete Ionic Equation – All soluble electrolytes shown as ions Pb +2 (aq) + 2NO 3 - (aq) + 2K + (aq) + 2I - (aq) PbI 2 (s) + 2K + (aq) + 2NO 3 - (aq) Net Ionic Equation Net Ionic Equation – shows only the ions and molecules directly involved in the equation Pb +2 (aq) + 2I - (aq) PbI 2 (s)

22 Writing Complete Ionic Equations 1. Start with a balanced molecular equation. 2. Break all soluble strong electrolytes (compounds with (aq) beside them) into their ions. indicate the correct formula and charge of each ion indicate the correct number of each ion write (aq) after each ion 3. Bring down all compounds with (s), (l), or (g) unchanged.

23 Writing Complete Ionic Equations Example: 2Na 3 PO 4 (aq) + 3CaCl 2 (aq) 6NaCl(aq) + Ca 3 (PO 4 ) 2 (s) Becomes… 6Na + (aq) + 2PO 4 3- (aq) + 3Ca 2+ (aq) + 6Cl - (aq) 6Na + (aq) + 6Cl - (aq) + Ca 3 (PO 4 ) 2 (s)

24 Spectator Ions Appear in identical forms among both the reactants and products of a complete ionic equation net ionic equations When writing net ionic equations they cancel each other out Pb +2 (aq) + 2NO 3 - (aq) + 2K + (aq) + 2I - (aq) PbI 2 (s) + 2K + (aq) + 2NO 3 - (aq)

25 Writing Net Ionic Equations Cancel out spectator ions from complete ionic equation then write whats left 6Na + (aq) + 2PO 4 3- (aq) + 3Ca 2+ (aq) + 6Cl - (aq) 6Na + (aq) + 6Cl - (aq) + Ca 3 (PO 4 ) 2 (s) Becomes… 2PO 4 3- (aq) + 3Ca 2+ (aq) Ca 3 (PO 4 ) 2 (s)

26 Practice Write complete ionic and net ionic equations for the following: 1. 3(NH 4 ) 2 CO 3 (aq) + 2Al(NO 3 ) 3 (aq) 6NH 4 NO 3 (aq) + Al 2 (CO 3 ) 3 (s) 2. 2NaOH(aq) + H 2 SO 4 (aq) Na 2 SO 4 (aq) + 2H 2 O(l) 3. Zn(s) + CuSO 4 (aq) --> ZnSO 4 (aq) + Cu(s)

27 Answers 1. Complete Ionic Equation: 6NH 4 + (aq) + 3CO 3 2- (aq) + 2Al 3+ (aq) + 6NO 3 - (aq) 6NH 4 + (aq) + 6NO 3 - (aq) + Al 2 (CO 3 ) 3 (s) Net Ionic Equation:2 Al 3+ (aq) + 3 CO 3 2- (aq) Al 2 (CO 3 ) 3 (s) 2. Complete Ionic Equation: 2Na + (aq) + 2OH - (aq) + 2H + (aq) + SO 4 2- (aq) 2Na + (aq) + SO 4 2- (aq) + 2H 2 O(l) Net Ionic Equation: OH - (aq) + H + (aq) H 2 O(l) *Note: simplify net ionic equations if possible 3. Complete Ionic Equation: Zn(s) + Cu 2+ (aq) + SO 4 2- (aq) Zn 2+ (aq) + SO 4 2- (aq) + Cu(s) Net Ionic Equation:Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s)

28 Stoichiometry The study of the quantitative aspects of chemical reactions.

29 Mole Ratio Conversion factor that relates amount in moles of any two substances involved in a chemical reaction 2 Al 2 O 3 (l) 4 Al(s) + 3 O 2 (g) Mole ratio Al 2 O 3 to O 2 = 2:3 Mole ratio Al to Al 2 O 3 = 4:2 or 2:1 Mole ratio Al to O 2 = 4:3

30 Stoichiometry Problems Solved just like conversions! You must start with a balanced chemical equation Types: Mole Mole Mass Mass Mass Mole or Mole Mass

31 Mole How many moles of O 2 are produced from 3.5 moles of Al 2 O 3 ? 3.5 mol Al 2 O 3 × = 5.25 mol O 2 *Use mole ratio to convert between moles! 2 Al 2 O 3 (l) 4 Al(s) + 3 O 2 (g) 3 mol O 2 2 mol Al 2 O 3

32 Mass How many grams of Al are produced from 4.56 grams of Al 2 O 3 ? Molar Mass Al 2 O 3 = g/molMolar Mass Al = g/mol 4.56 g Al 2 O 3 × × × = 2.41 g Al 1 mol Al 2 O g Al 2 O 3 4 mol Al 2 mol Al 2 O g Al 1 mol Al 2 Al 2 O 3 (l) 4 Al(s) + 3 O 2 (g)

33 Limiting/Excess Reactant Recipe makes 10 pancakes 3 eggs 2 cups bisquik 1 cup milk 1 cup chocolate chips What is the most amount of pancakes that I can make with 6 eggs and 5 cups of milk? What is the most amount of pancakes that I can make with 3 cups of chocolate chips and 8 cups of milk? What limits how many pancakes I can make and what will be left over?

34 Limiting/Excess Reactant The limiting reactant is the reactant that is consumed first, limiting the amounts of products formed. The excess reactant is the reactant that is leftover after the reaction has gone to completion.

35 ReactantsProducts 2 NO(g) + O 2 (g) 2 NO 2 (g) Limiting reactant = ___________ Excess reactant = ____________ Limiting/Excess Reactant

36 Calculating Limiting/Excess Reagent Given 12.4 grams of NO and 9.40 grams of O 2, which is the limiting and which is the excess reagent? 2 NO(g) + O 2 (g) 2 NO 2 (g) 1 mol NO g NO 2 mol NO 2 2 mol NO g NO 2 1 mol NO g NO × × × = g NO 2 1 mol O g O 2 2 mol NO 2 1 mol O g NO 2 1 mol NO g O 2 × × × = g NO 2

37 Calculating Limiting/Excess Reagent NO limits the amount of NO 2 that is made Limiting reagent = NO O 2 will be leftover once the reaction is complete Excess reactant = O 2 2 NO(g) + O 2 (g) 2 NO 2 (g) 1 mol NO g NO 2 mol NO 2 2 mol NO g NO 2 1 mol NO g NO × × × = g NO 2 1 mol O g O 2 2 mol NO 2 1 mol O g NO 2 1 mol NO g O 2 × × × = g NO 2

38 Calculating Limiting/Excess Reagent How much O 2 will be in excess once the reaction is complete? 2 NO(g) + O 2 (g) 2 NO 2 (g) 1 mol NO g NO 1 mol O 2 2 mol NO g O 2 1 mol O g NO × × × = 6.61 g O grams of O 2 will be used in the reaction. You have 9.40 grams to start with – 6.61 = 2.79 grams O 2 in excess (leftover)

39 Limiting/Excess Reactant If the equation has 2 or more products, when determining the limiting/excess reactants, simply pick one of the products and convert both reactants to that product. You MUST use the same product for both.

40 Percent Yield Percentage comparing how much product was actually produced compared to what should have been produced. Calculate theoretical yield using stoichiometry. If you know how much of each reactant you start out with, use stoichiometry to calculate how much of the given product you should produce. Actual Yield Theoretical Yield × 100

41 Percent Yield An experiment was performed combining using 3.4 g of AgNO 3 and an unlimited supply of KCl. If the experiment yielded 2.7 g of AgCl, what is the percent yield of the experiment? AgNO 3 (aq) + KCl(aq) AgCl(s) + KNO 3 (aq) 1 mol AgNO g AgNO 3 1 mol AgCl 1 mol AgNO g AgCl 1 mol AgCl 3.4 g AgNO 3 × × × = 2.9 g AgCl Percent Yield = × 100 = 93%


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