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Chemical Reactions

Learn how to balance chemical equations and understand the stoichiometry of reactions. Explore different reaction types such as synthesis, decomposition, combustion, single replacement, and double replacement. Discover the mole ratios and calculations involved in stoichiometric calculations.

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Chemical Reactions

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  1. ChemicalReactions

  2. Chemical Equations • A chemical reaction shows the formulas and relative amounts of reactants and products in a reaction. 4 Al (s) + 3 O2 (g)  2 Al2O3 (s) reactants product stoichiometric coefficients physical state s, l, g or aq

  3. Chemical Equations • 4 Al (s) + 3 O2 (g)  2 Al2O3 (s) • Interpret this equation as • 4 atoms solid Al react with 3 molecules gaseous O2 to form 2 formula units of solid Al2O3 • 4 moles solid Al react with 3 moles gaseous O2 to form 2 moles of solid Al2O3

  4. Balancing Chemical Equations “Matter is conserved in chemical change” Antoine Lavoisier, 1789 An equation must be balanced: It must have the same number of atoms of each kind on both sides

  5. The rules of the game • Write the correct formulas of the reactants and products • Do not change the formulas to balance the equation • Put a coefficient in front of each formula so that the same number of atoms of each kind appear in both the reactants and products • The coefficent multiplies through its formula • 2 H2O shows 4 H atoms and 2 O atoms

  6. Example 1 • Balance this equation: N2 + H2 NH3 • There are 2 Ns on the left, we need 2 Ns on the right • Put a coefficient of 2 in front of NH3 • N2 + H2 2 NH3 • Now there are 6 Hs on the right, we need 6 on the left • Put a coefficient of 3 in front of H2 • N2 + 3 H2 2 NH3 • Balanced equation shows 2 Ns and 6 Hs on each side.

  7. Example 2 • Balance this equation: CH4 + O2 CO2 + H2O • There is one C on each side. Cs are balanced • There are 4 Hs on left, 2 on right. Put a 2 in front of H2O • CH4 + O2 CO2 + 2 H2O • There are 2 + 2 = 4 Os on right. Put a 2 in front of O2 • CH4 + 2 O2 CO2 + 2 H2O • Balanced equation has 1 C, 4 Hs, and 4 Os on each side

  8. Balance these equationsand list the coefficients from left to right(including coefficients of one) • HgO (s)  Hg (l) + O2 (g) • 2 HgO (s)  2 Hg (l) + O2 (g) Answer 2, 2, 1 • Mg (s) + HCl (aq)  MgCl2 (aq) + H2 (g) • Mg (s) + 2 HCl (aq)  MgCl2 (aq) + H2 (g) Answer 1, 2, 1, 1

  9. Balance these equationsand list the coefficients from left to right(including coefficients of one) • C3H8 (g) + O2 (g)  CO2 (g) + H2O (l) • C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l) Answer 1, 5, 3, 4 • Al (s) + Cl2 (g)  AlCl3 (s) • 2 Al (s) + 3 Cl2 (g)  2 AlCl3 (s) Answer 2, 3, 2

  10. Balance these equationsand list the coefficients from left to right(including coefficients of one) • H2O2 (aq)  H2O (l) + O2 (g) • 2 H2O2 (aq)   H2O (l) + O2 (g) Answer 2, 2, 1 • C4H10 (g) + O2 (g)  CO2 (g) + H2O (l) • 2 C4H10 (g) + 13 O2 (g)  8 CO2 (g) + 10 H2O (l) Answer 2, 13, 8, 10

  11. Combustion Reactions • In combustion, a hydrocarbon or C–H–O fuel combines with O2 to form CO2 and H2O __ CH4 + __ O2 __ CO2 + __ H2O 1 CH4 + 2 O2 1 CO2 + 2 H2O Balanced equation shows 1 C, 4 H, and 4 O on each side • If N or S are in the formula for the fuel, assume it is oxidized to NO2 or SO2

  12. Reaction types • Synthesis or combination A + B → C • 2 or more formulas combine to make one product • Decomposition C → A + B • One formula breaks apart into 2 or more formulas • Combustion • C, H, O fuel + O2→ CO2 + H2O • Single replacement A + BX → AX + B • One element pushes another element out of a compound • Double replacement AX + BY → AY + BX • Two compounds switch partners

  13. Reaction types • Classify each reaction as combination, decomposition, combustion, single replacement, or double replacement: • 2 C2H6 + 7 O2→ 4 CO2 + 6 H2O • 2 Na + Cl2 → 2 NaCl • Cl2 + 2 KBr → 2 KCl + Br2 • NH4NO3 → N2O + 2 H2O • K2SO4 + BaCl2 → BaSO4 + 2 KCl

  14. Stoichiometry • Stoichiometry is chemical accounting • Heart of stoichiometry is mole ratio given by coefficients of balanced equation

  15. mole ratio moles B moles A moles A moles B Stoichiometry • Stoichiometry is chemical accounting • Heart of stoichiometry is mole ratio given by coefficients of balanced equation

  16. Example • How many moles of O2 are produced from the decomposition of 1.76 mol KClO3: 2 KClO3 (s)  2 KCl (s) + 3 O2 (g) • Question about O2 but information for KClO3 • O2 and KClO3 related by coefficients of balanced equation. • Coefficients are in moles: 2 mol KClO3 : 3 mol O2

  17. Example • How many moles of O2 are produced from the decomposition of 1.76 mol KClO3: 2 KClO3 (s)  2 KCl (s) + 3 O2 (g)

  18. Example • How many moles of CO2 form when 0.35 mol C2H6 burn? 2 C2H6 + 7 O2→ 4 CO2 + 6 H2O

  19. Example • How many moles of CO2 form when 0.35 mol C2H6 burn? 2 C2H6 + 7 O2→ 4 CO2 + 6 H2O

  20. Example • How many grams of Al react with 0.15 moles HCl: 2 Al (s) + 6 HCl (aq)  2 AlCl3 (aq) + 3 H2 (g) • Coefficients are in moles • Two steps: convert mol HCl to mol Al, then convert mol Al to grams

  21. Example • How many grams of Al react with 0.15 moles HCl: 2 Al (s) + 6 HCl (aq)  2 AlCl3 (aq) + 3 H2 (g) • Coefficients are in moles • Two steps: convert mol HCl to mol Al, then convert mol Al to grams

  22. Example • How many moles of Ag are produced when 10.0 g Ag2O decompose: 2 Ag2O (s)  4 Ag (s) + O2 (g) • Two steps: first convert g Ag2O to mol Ag2O, then use mole ratio to find moles Ag

  23. Example • How many moles of Ag are produced when 10.0 g Ag2O decompose: 2 Ag2O (s)  4 Ag (s) + O2 (g) • Two steps: first convert g Ag2O to mol Ag2O, then use mole ratio to find moles Ag

  24. Example • How many grams of H2 (g) are needed to produce 100.0 g of CH3OH: CO (g) + 2 H2 (g)  CH3OH (l)

  25. Example • How many grams of O2 react with 268 grams of octane (C8H18) in the combustion of octane? 2 C8H18 + 25 O2 16 CO2 + 18 H2O

  26. Example • How many grams of O2 react with 268 grams of octane (C8H18) in the combustion of octane? 2 C8H18 + 25 O2 16 CO2 + 18 H2O

  27. mole ratio moles B moles A moles A moles B grams A grams B Stoichiometry • Convert to moles (÷ molar mass) • Use the mole ratio • Convert new moles back to grams (x new molar mass)

  28. Example • Balance the chemical equation • Identify the type of reaction • Calculate how many moles of AgNO3 are needed to produce 3.17 g Ag2CO3 AgNO3 + K2CO3 Ag2CO3 + KNO3

  29. Example • Balance the chemical equation • Identify the type of reaction • Calculate how many grams of Al are needed to produce 0.045 g H2 Al (s) + HCl (aq)  AlCl3 (aq) + H2 (g)

  30. Chemical Reactions in Solution • Most reactions occur in aqueous solution • SOLUTE is the substance to be dissolved in solution • SOLVENT is the substance (often a liquid) the solute dissolves in • The concentration of the solution is Molarity (M) = moles solute L solution

  31. Example 4-8B • 15.0 mL of concentrated acetic acid, HC2H3O2 (d = 1.048 g/mL), are dissolved in enough water to produce 500.0 mL of solution. What is the concentration of the solution?

  32. Example 4-9B • How many grams of Na2SO4 • 10 H2O are needed to prepare 355 mL of 0.445 M Na2SO4?

  33. Dilution problems • It is common to prepare a solution by diluting a more concentrated solution (the stock solution). • The moles of solute taken from the stock solution are given by moles solute = volume x molarity • All the solute taken from the stock appears in the diluted solution, so moles solute are constant: VstockMstock = VdiluteMdilute

  34. Example 4-10A • 15.00 mL of 0.450 M K2CrO4 solution are diluted to 100.00 mL. What is the concentration of the dilute solution?

  35. Example 4-10B • After being left out in an open beaker, 275 mL of 0.105 M NaCl has evaporated to only 237 mL. What is the concentration of the solution after evaporation?

  36. grams A grams B mole ratio moles B moles A moles A moles B mL A mL B Stoichiometry in Solution • Stoichiometry in solution is just the same as for mass problems, except the conversion into or out of moles uses molarity instead of molar mass:

  37. Example 4-11B K2CrO4 (aq) + 2 AgNO3 (aq)  Ag2CrO4 (s) + 2 KNO3 (aq) • How many mL of 0.150 M AgNO3 must react with excess K2CrO4 to produce exactly 1.00 g Ag2CrO4?

  38. Limiting reactant • In a given reaction, often there is not enough of one reactant to use up the other reactant completely • The reactant in short supply LIMITS the quantity of product that can be formed

  39. Goldilocks Chemistry • Imagine reacting different amounts of Zn with 0.100 mol HCl: Zn (s) + 2 HCl (aq)  ZnCl2 (aq) + H2 (g) Rxn 1Rxn 2Rxn 3 Mass Zn 6.54 g3.27 g1.31 g Moles Zn 0.100 mol0.0500 mol0.0200 mol Moles HCl 0.100 mol0.100 mol0.100 mol Ratio mol HCl1.002.005.00 mol Zn

  40. Limiting reactant problems • The easiest way to do these is to do two stoichiometry calculations • Find the amount of product possible from each reactant • The smaller answer is the amount of product you can actually make (you just ran out of one reactant) • The reactant on which that answer was based is the limiting reactant

  41. Example 4-13B • 12.2 g H2 and 154 g O2 are allowed to react. Identify the limiting reactant, which gas remains after the reaction, and what mass of it is left over. 2 H2 (g) + O2 (g)  2 H2O (l)

  42. Percent Yield • In real experiments we often do not get the amount of product we calculate we should, because • the reactants may participate in other reactions (side reactions) that produce other products (by-products) • The reaction often does not go to completion. • Percent yield tells the ratio of actual to theoretical amount formed.

  43. Percent Yield • Suppose you calculate that a reaction will produce 50.0 g of product. This is the theoretical yield. • The reaction actually produces only 45.0 g of product . This is the actual yield. • Percent yield = 45.0 g (actual) x 100 = 90.0% 50.0 g (theoretical)

  44. Example 4-14B • What is the percent yield if 25.0 g P4 reacts with 91.5 g Cl2 to produce 104 g PCl3: P4 (s) + 6 Cl2 (g)  4 PCl3 (l)

  45. Example 4-15B • What mass of C6H11OH should you start with to produce 45.0 g C6H10 if the reaction has 86.2% yield and the C6H11OH is 92.3% pure: C6H11OH (l)  C6H10 + H2O (l)

  46. Exercise 26 • Balance these equations by inspection • (NH4)2Cr2O7 (s)  Cr2O3 (s) + N2 (g) + H2O (g) • NO2 (g) + H2O (l)  HNO3 (aq) + NO (g) • H2S (g) + SO2 (g)  S (g) + H2O (g) • SO2Cl2 + HI  H2S + H2O + HCl + I2

  47. Exercise 30 • Write balanced equations for these reactions: • Sulfur dioxide gas with oxygen gas to produce sulfur trioxide gas • Solid calcium carbonate with water and dissolved carbon dioxide to produce aqueous calcium hydrogen carbonate • Ammonia gas and nitrogen monoxide gas to produce nitrogen gas and water vapor

  48. Exercise 32 • 3 Fe (s) + 4 H2O (g)  Fe3O4 (s) + H2 (g) • How many moles of H2 can be produced from 42.7 g Fe and excess steam? • How many grams of H2O are consumed in the conversion of 63.5 g Fe to Fe3O4? • If 7.36 mol H2 are produced, how many grams of Fe3O4 must also be produced?

  49. Exercise 36 • Silver oxide decomposes above 300 °C to yield metallic silver and oxygen gas. 3.13 g impure silver oxide yields 0.187 g O2. Assuming there is no other source of O2, what is the % Ag2O by mass in the original sample?

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