1 / 54

Chapter 6

Chapter 6. Chemical Reactions: Classification and Mass Relationships. Balancing Chemical Equations. Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations (chemical reactions) reactants  products limestone  quicklime + gas

rosiee
Download Presentation

Chapter 6

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 6 Chemical Reactions: Classification and Mass Relationships

  2. Balancing Chemical Equations • Alphabet – elemental symbols • Words – chemical formulas • Sentences – chemical equations (chemical reactions) reactants  products limestone  quicklime + gas Calcium carbonate  calcium oxide + carbon dioxide CaCO3(s)  CaO(s) + CO2(g)

  3. Balancing Chemical Equations • Chemical reactions include • Reactants • Products • Balanced – Law of Conservation of Mass • # of atoms of an element on the reactant side must equal the # of atoms of that element on the product side. • Indicate the state of matter of each chemical in the reaction (Chapter 4)

  4. Balancing Chemical Equations • Write the equation without coefficients • List the elements in each equation • Secret: if the same polyatomic ion exists on both sides, keep it together • Determine the # of each kind of atom on both sides • Balance atoms one element at a time by adjusting coefficients • DO NOT ALTER THE FORMULA OF THE COMPOUND!!!!! • Only coefficients can be altered • Secret: • Balance atoms appearing only once on each side first. • Save compounds comprised of only one type of element till last. • Reduce to lowest terms if necessary

  5. Examples • Balance the following equations: • Al(s) + Fe2O3(s) → Al2O3 (s) + Fe (l) • Solid copper reacts with aqueous silver nitrate to form aqueous copper (II) nitrate and silver solid • H3PO4 (l) → H2O (l) + P4O10 (s) • C4H10(g) + O2 (g) → CO2(g) + H2O (g)

  6. Avogadro’s Number and the Mole • Meaning of a chemical reaction 2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O (g) • 2 molecules of C4H10(g) reacts with 13 molecules of O2(g) to form 8 molecule of CO2(g) and 10 molecules of H2O(g)

  7. Avogadro’s Number and the Mole • Molecule’s mass = the sum of the atomic masses of the atoms making up the molecule. • m(C2H4O2) = 2·mC + 4·mH + 2·mO • = 2·(12.01) + 4·(1.01) + 2·(16.00) • = 60.06 amu

  8. Avogadro’s Number of the Mole • One mole (mol) of any substance contains 6.02 x 1023 (Avogadro’s Number) units of that substance. • One mole (mol) of a substance is the gram mass value equal to the amu mass of the substance. • Calculated the same as amu’s for a molecule

  9. Avogadro’s Number and the Mole • Calculate the molar mass of the following: • Fe2O3 (Rust) • C6H8O7 (Citric acid) • C16H18N2O4 (Penicillin G)

  10. Avogadro’s Number and the Mole • Methionine, an amino acid used by organisms to make proteins, is represented below. Write the formula for methionine and calculate its molar mass. (red = O; gray = C; blue = N; yellow = S; ivory = H)

  11. Stoichiometry • 4 Conversion units • Chemical formula • Balanced chemical equation • Coefficients can read as; • # of molecules • # of moles of that molecules • Allows conversion between compounds in an equation • Avogadro’s # - 6.02 x 1023 of X = 1 mole of X • Molar mass – how many grams of a substance = 1 mole of that substance

  12. Stoichiometric Calculations

  13. Avogadro’s Number and the Mole • How many grams of oxygen are present in 5.961 x 1020 molecules of KClO3? How many atoms of oxygen are present?

  14. Avogadro’s Number and the Mole • Calculate the number of oxygen atoms in 29.34 g of sodium sulfate, Na2SO4.  • A.  1.244 × 1023 O atoms • B.  4.976 × 1023 O atoms • C.  2.409 × 1024 O atoms • D.  2.915 × 1024 O atoms • E.  1.166 × 1025 O atoms

  15. Problem • Potassium dichromate, K2Cr2O7, is used in tanning leather, decorating porcelain and water proofing fabrics. Calculate the number of chromium atoms in 78.82 g of K2Cr2O7.  • A.  9.490 × 1025 Cr atoms • B.  2.248 × 1024 Cr atoms • C.  1.124 × 1024 Cr atoms • D.  3.227 × 1023 Cr atoms • E.  1.613 × 1023 Cr atoms

  16. Stoichiometry: Chemical Arithmetic

  17. Stoichiometry: Equation Arithmetic • Balance the following, and determine how many moles of CO will react with 0.500 moles of Fe2O3. Fe2O3(s) + CO(g) → Fe(s) + CO2(g)

  18. Stoichiometry: Chemical Arithmetic • Aqueous sodium hydroxide and chlorine gas are combined to form aqueous sodium hypochlorite (household bleach), aqueous sodium chloride and liquid water. • How many grams of NaOH are needed to react with 25.0 g of Cl2?

  19. Problem • Sulfur dioxide reacts with chlorine to produce thionyl chloride (used as a drying agent for inorganic halides) and dichlorine monoxide (used as a bleach for wood, pulp and textiles).SO2(g) + 2Cl2(g) → SOCl2(g) + Cl2O(g)If 0.400 mol of Cl2 reacts with excess SO2, how many moles of Cl2O are formed?  • A.  0.800 mol • B.  0.400 mol • C.  0.200 mol • D.  0.100 mol • E.  0.0500 mol

  20. Problem • Nitrogen gas and hydrogen gas are combined to form ammonia (NH3), an important source of fixed nitrogen that can be metabolized by plants, using the Haber process.How many grams of nitrogen are needed to produce 325 grams of ammonia?  • A.  1070 g • B.  535 g • C.  267 g • D.  178 g • E.  108 g

  21. Percent Yields • Yields of Chemical Reactions:If the actual amount of product formed in a reaction is less than the theoretical amount, we can calculate apercentage yield.

  22. Problem • What is the percent yield for the reactionPCl3(g) + Cl2(g) → PCl5(g)if 119.3 g of PCl5 ( MM = 208.2 g/mol) are formed when 61.3 g of Cl2 (  MM = 70.91 g/mol) react with excess PCl3?  • A.  195% • B.  85.0% • C.  66.3% • D.  51.4% • E.  43.7%

  23. Types of Chemical Reactions • Chemical Reactions discussed in College Chemistry can be broken down into 3 main categories • Precipitation reactions • Acid-Base reactions • Oxidation-Reduction (redox) reactions

  24. Types of Chemical Reactions • Precipitation Reactions:A process in which an insoluble solid (precipitate) drops out of the solution. • Clear solutions of two ionic compounds when mixed form a cloudy solution (cloudiness indicates solid)

  25. Types of Reactions • Acid–Base Neutralization:A process in which an acid reacts with a base to yield water plus an ionic compound called a salt. • The driving force of this reaction is the formation of the stable water molecule.

  26. Types of Reaction • Metathesis Reactions (Double Displacement Reaction): These are reactions where two reactants just exchange parts. AX + BY AY + BX

  27. Types of Reactions • Oxidation–Reduction (Redox) Reaction:A process in which one or more electrons are transferred between reaction partners. • The driving force of this reaction is the decrease in electrical potential.

  28. Precipitation Reactions • Develop the reaction equation • Balance the reaction equation • Predict the state of matter of each species present

  29. Precipitation Reactions and Solubility Rules • To predict whether a precipitation reaction will occur must be able to predict whether a compound is soluble or not • Solubility rules

  30. Solubility Rules • Salts - soluble: • All alkali metal and ammonium ion salts • All salts of the NO3–, ClO3–, ClO4–, C2H3O2–,and HCO3–ions

  31. Solubility Rules • Salts which are soluble with exceptions: • Cl–, Br–, I– ion salts except with Ag+, Pb2+, & Hg22+ • SO42– ion salts except with Ag+, Pb2+, Hg22+, Ca2+, Sr2+, & Ba2+

  32. Solubility Rules • Salts which are insoluble with exceptions: • O2– & OH– ion salts except with the alkali metal ions, and Ca2+, Sr2+, & Ba2+ ions • CO32–, PO43–, S2–, CrO42–, & SO32– ion salts except with the alkali metal ions and the ammonium ion

  33. Precipitation Reactions and Solubility Rules • Predict the solubility of: • (a) CdCO3(b) MgO (c) Na2S (d) PbSO4(e) (NH4)3PO4(f) HgCl2

  34. Precipitation Reaction • Precipitation reactions only occur if a solid is produced as a product. • If all products are aqueous compounds then no reaction has taken place.

  35. Precipitation Reactions and Solubility Guidelines • Predict whether a precipitate will form for: • (a) NiCl2(aq) + (NH4)2S(aq)  • (b) Na2CrO4(aq) + Pb(NO3)2(aq) • (c) AgClO4(aq) + CaBr2(aq)

  36. Problem • Select the precipitate that forms when aqueous ammonium sulfide reacts with aqueous copper(II) nitrate.  • A.  CuS • B.  Cu2S • C.  NH4NO3 • D.  NH4(NO3)2 • E.  CuSO4

  37. Problem • Select the precipitate that forms when the following reactants are mixed.Mg(CH3COO)2(aq) + LiOH(aq) →  • A.  LiCH3COO • B.  Li(CH3COO)2 • C.  MgOH • D.  Mg(OH)2 • E.  CH3OH

  38. Acids, Bases and Neutralization Reactions • Acid / Base Definitions • Arrhenius • Acid – donates a H+ (H3O+) • Base – donates an OH- • Bronsted-Lowry • Acid – donates a H+ • Base – H+ acceptor

  39. Acids, Bases and Neutralization Reactions

  40. Acids, Bases and Neutralization Reactions • Neutralization Reaction:produces salt & water. • HA(aq) + MOH(aq)  H2O(l) + MA(aq) • Write a balanced chemical equation for the following: • (a) HBr(aq) + Ba(OH)2(aq) • (b) HCl(aq) + NH3(aq)

  41. Oxidation-Reduction Reactions • Redox reactions are those involving the oxidation and reduction of species (element or ion of an element). • Oxidation and reduction mustoccur together. They cannot exist alone. • Two important types • Single displacement reactions (activity series) • Combustions – reaction of a substance with O2

  42. Oxidation Reduction Reactions Oxidation Is Loss (of electrons) AnodeOxidation Reducing Agent

  43. Oxidation Reduction Reactions Reduction Is Gain (of electrons) Cathode Reduction Oxidizing Agent

  44. Redox Reactions • Assigning Oxidation Numbers:All atoms have an “oxidation number” regardless of whether it carries an ionic charge. 1. An atom in its elemental state has an oxidation number of zero. 2. An atom in a monatomic ion has an oxidation number identical to its charge.

  45. Redox Reactions 3. An atom in a polyatomic ion or in a molecular compound usually has the same oxidation number it would have if it were a monatomic ion. • A. Hydrogen can be either +1 or –1. • B. Oxygen usually has an oxidation number of –2. • In peroxides, oxygen is –1. • C. Halogens usually have an oxidation number of –1. • When bonded to oxygen, chlorine, bromine, and iodine have positive oxidation numbers.

  46. Redox Reactions 4. The sum of the oxidation numbers must be zero for a neutral compound and must be equal to the net charge for a polyatomic ion. • A. H2SO4 2(+1) + (?) + 4(–2) = 0 net charge ? = 0 – 2(+1) – 4(–2) = +6 • B. ClO4– (?) + 4(–2) = –1 net charge ? = –1 – 4(–2) = +7

  47. Problem • Sodium tripolyphosphate is used in detergents to make them effective in hard water. Calculate the oxidation number of phosphorus in Na5P3O10.  • A.  +3 • B.  +5 • C.  +10 • D.  +15 • E.  none of these is the correct oxidation number

  48. Problem • The oxidation numbers of P, S and Cl in H2PO2-, H2S and KClO4 are, respectively  • A.  -1, -1, +3 • B.  +1, -2, +7 • C.  +1, +2, +7 • D.  -1, -2, +7 • E.  -1, -2, +3

  49. Redox Reactions • Whenever one atom loses electrons (is oxidized), another atom must gain those electrons (be reduced). • A substance which loses electrons (oxidized) is called a reducing agent. Its oxidation number increases. • Asubstance which gains electrons (reduced) is called the oxidizing agent. Its oxidation number decreases.

  50. Redox Reactions • For each of the following, identify which species is the reducing agent and which is the oxidizing agent. • Ca(s) + 2 H+(aq)  Ca2+(aq) + H2(g) • 2 Fe2+(aq) + Cl2(aq) 2 Fe3+(aq) + 2 Cl–(aq) • SnO2(s) + 2 C(s)  Sn(s) + 2 CO(g) • Sn2+(aq) + 2 Fe3+(aq)  Sn4+(aq) + 2 Fe2+(aq)

More Related