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Reactions & Stoichiometry

UNIT 6. Reactions & Stoichiometry. Overview. Reactions Write formula/word equations Balance Equations Identify Types Predict Products Write Net Ionic Equations Stoichiometry Conversions Limiting & Excess Reagent Percent Yield. Chemical Reactions.

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Reactions & Stoichiometry

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  1. UNIT 6 Reactions & Stoichiometry

  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 Reactants: Zn + I2 Product: Zn I2

  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 O2(g) ---> 2 Al2O3(s) • Reactants react to produce products • The letters (s), (g), (l), and (aq) are the physical states of compounds. • “aq” represents aqueous meaning dissolved in water (solution) • The numbers in the front are called coefficients. • Subscripts represent the number of each atom in a compound (Reactants) (Products)

  6. Chemical Reactions

  7. Diatomic Elements • Elements that cannot exist by themselves (always occur in pairs) • Bromine (Br2) • Iodine (I2) • Nitrogen (N2) • Chlorine (Cl2) • Hydrogen (H2) • Oxygen (O2) • Fluorine (F2) BrINClHOF!

  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)+ H2SO4(aq)  Li2SO4(aq) + 2 H2O(l) 2. When crystalline C6H12O6 is burned in oxygen, carbon dioxide and water vapor are formed. C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(g)

  9. Balancing Equations • 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 2 Hg atoms, 2 O atoms 2HgO(s) ---> 2 Hg(l) + O2(g)

  10. Balancing Equations ___ Al(s) + ___ Br2(l) ---> ___ Al2Br6(s) __C3H8(g) + __ O2(g)  __ CO2(g) + __ H2O(g) __ B4H10(g) + __ O2(g)  __ B2O3(g) + __ H2O(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 H2+ O2 H2O • Metal carbonates CaO+ CO2 CaCO3 • Metal hydroxides CaO+ H2O Ca(OH)2 • Metal chlorates KCl+ O2 KClO3 • Oxyacids CO2+ H2O  H2CO3

  13. Decomposition Reactions • A single compound breaks down into two or more simpler substances AX  A + X • Decomposition of: • Binary compounds H2O  H2+ O2 • Metal carbonates CaCO3 CaO+ CO2 • Metal hydroxides Ca(OH)2 CaO+ H2O • Metal chlorates KClO3 KCl+ O2 • Oxyacids H2CO3 CO2+ H2O

  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 • One of the compounds formed is usually a precipitate, an insolublegas that bubbles out of solution, or a molecularcompound, usually water.

  17. 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 – 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

  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 CxHx + O2 CO2 + H2O

  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 + H2O

  21. Chemical Equations • Molecular Equation– shows complete chemical formulas of reactants and products Pb(NO3)2(aq) + 2KI(aq)  PbI2(s) + 2KNO3(aq) • Complete Ionic Equation – All soluble electrolytes shown as ions Pb+2(aq) + 2NO3-(aq) + 2K+(aq) + 2I-(aq)  PbI2(s) + 2K+(aq) + 2NO3-(aq) • Net Ionic Equation – shows only the ions and molecules directly involved in the equation Pb+2(aq) + 2I-(aq)  PbI2(s)

  22. Writing Complete Ionic Equations • Start with a balanced molecular equation. • 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 • Bring down all compounds with (s), (l), or (g) unchanged.

  23. Writing Complete Ionic Equations Example: 2Na3PO4(aq) + 3CaCl2(aq)  6NaCl(aq) + Ca3(PO4)2(s) Becomes… 6Na+(aq) + 2PO43-(aq) + 3Ca2+(aq) + 6Cl-(aq)  6Na+(aq) + 6Cl-(aq) + Ca3(PO4)2(s)

  24. Spectator Ions • Appear in identical forms among both the reactants and products of a complete ionic equation • When writing net ionic equationsthey cancel each other out Pb+2(aq) + 2NO3-(aq) + 2K+(aq) + 2I-(aq)  PbI2(s) + 2K+(aq) + 2NO3-(aq)

  25. Writing Net Ionic Equations • Cancel out spectator ions from complete ionic equation then write what’s left 6Na+(aq)+ 2PO43-(aq) + 3Ca2+(aq) + 6Cl-(aq)6Na+(aq) + 6Cl-(aq) + Ca3(PO4)2(s) Becomes… 2PO43-(aq) + 3Ca2+(aq)  Ca3(PO4)2(s)

  26. Practice Write complete ionic and net ionic equations for the following: • 3(NH4)2CO3(aq) + 2Al(NO3)3(aq)  6NH4NO3(aq) + Al2(CO3)3(s) • 2NaOH(aq) + H2SO4(aq)  Na2SO4(aq) + 2H2O(l) • Zn(s) + CuSO4(aq) --> ZnSO4(aq) + Cu(s)

  27. Answers • Complete Ionic Equation: 6NH4+(aq) + 3CO32-(aq) + 2Al3+(aq) + 6NO3-(aq)6NH4+(aq) + 6NO3-(aq) + Al2(CO3)3(s) Net Ionic Equation: 2 Al3+(aq) + 3 CO32-(aq)  Al2(CO3)3(s) • Complete Ionic Equation: 2Na+(aq) + 2OH-(aq) + 2H+(aq) + SO42-(aq) 2Na+(aq) + SO42-(aq) + 2H2O(l) Net Ionic Equation: OH-(aq) + H+(aq)  H2O(l) *Note: simplify net ionic equations if possible • Complete Ionic Equation: Zn(s) + Cu2+(aq) + SO42-(aq)  Zn2+(aq) + SO42-(aq) + Cu(s) Net Ionic Equation: Zn(s) + Cu2+(aq)  Zn2+(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 Al2O3(l)  4 Al(s) + 3 O2(g) Mole ratio Al2O3 to O2 = 2:3 Mole ratio Al to Al2O3 = 4:2 or 2:1 Mole ratio Al to O2 = 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  Mole 2 Al2O3(l)  4 Al(s) + 3 O2(g) • How many moles of O2 are produced from 3.5 moles of Al2O3? 3.5 mol Al2O3 × = 5.25 mol O2 *Use mole ratio to convert between moles! 3 mol O2 2 mol Al2O3

  32. Mass  Mass 2 Al2O3(l)  4 Al(s) + 3 O2(g) • How many grams of Al are produced from 4.56 grams of Al2O3? • Molar Mass Al2O3 = 101.96 g/mol Molar Mass Al = 26.98 g/mol 4.56 g Al2O3 × × × = 2.41 g Al 1 mol Al2O3 101.96 g Al2O3 26.98 g Al 1 mol Al 4 mol Al 2 mol Al2O3

  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. 2 NO(g) + O2 (g) 2 NO2(g) Limiting/Excess Reactant Reactants Products Limiting reactant = ___________ Excess reactant = ____________

  36. 2 NO(g) + O2 (g) 2 NO2(g) Calculating Limiting/Excess Reagent • Given 12.4 grams of NO and 9.40 grams of O2, which is the limiting and which is the excess reagent? 1 mol NO 30.01 g NO 46.01 g NO2 1 mol NO2 2 mol NO2 2 mol NO 12.4 g NO × × × = 19.01 g NO2 1 mol O2 16.00 g O2 46.01 g NO2 1 mol NO2 2 mol NO2 1 mol O2 9.40 g O2 × × × = 54.06 g NO2

  37. 2 NO(g) + O2 (g) 2 NO2(g) Calculating Limiting/Excess Reagent • NO limits the amount of NO2 that is made • Limiting reagent = NO • O2 will be leftover once the reaction is complete • Excess reactant = O2 1 mol NO 30.01 g NO 46.01 g NO2 1 mol NO2 2 mol NO2 2 mol NO 12.4 g NO × × × = 19.01 g NO2 1 mol O2 32.00 g O2 46.01 g NO2 1 mol NO2 2 mol NO2 1 mol O2 9.40 g O2 × × × = 27.03 g NO2

  38. 2 NO(g) + O2 (g) 2 NO2(g) Calculating Limiting/Excess Reagent • How much O2 will be in excess once the reaction is complete? 1 mol NO 30.01 g NO 32.00 g O2 1 mol O2 1 mol O2 2 mol NO 12.4 g NO × × × = 6.61 g O2 • 6.61 grams of O2 will be used in the reaction. You have 9.40 grams to start with. • 9.40 – 6.61 = 2.79 grams O2 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 Actual Yield Theoretical Yield × 100 • 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.

  41. Percent Yield AgNO3(aq) + KCl(aq) → AgCl(s) + KNO3(aq) An experiment was performed combining using 3.4 g of AgNO3 and an unlimited supply of KCl. If the experiment yielded 2.7 g of AgCl, what is the percent yield of the experiment? 1 mol AgNO3 169.88 g AgNO3 143.32 g AgCl 1 mol AgCl 1 mol AgCl 1 mol AgNO3 3.4 g AgNO3 × × × = 2.9 g AgCl 2.7 2.9 Percent Yield = × 100 = 93%

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