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Chapter 4 Chemical Reactions PowerPoint Presentation
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Chapter 4 Chemical Reactions

Chapter 4 Chemical Reactions

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Chapter 4 Chemical Reactions

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  1. Chemistry B11 Chapter 4 Chemical Reactions

  2. Chemical Reactions Chemical change = Chemical reaction Substance(s) is used up (disappear) New substance(s) is formed. Different physical and chemical properties.

  3. Chemical Reactions

  4. Chemical Reactions A + B  C + D Reactants Products Chemical Equation

  5. AB  A + B 2. Decomposition (analysis) 2NaCl  2Na + Cl2 A + BC  AC + B 3. Single replacement reaction Fe + CuSO4 FeSO4 + Cu AB + CD  AD + CB 4. Double replacement reaction NaCl + AgNO3 NaNO3 + AgCl Chemical Reactions A + B  AB 1. Synthesis reaction (combination) 2H2 + O2 2H2O

  6. Solid (s) Liquid (l) Gas (g) Aqueous (aq) Ca(OH)2(s) + 2HCl(g)  CaCl2(s) + H2O(l) Chemical Reactions 5. AB + xO2 yCO2 + zH2O Combustion C3H8 + 5O2 3CO2 + 4H2O

  7. Balance a chemical equation Why balancing?

  8. Balance a chemical equation Law of conservation of mass Atoms are neither destroyed nor created. They shift from one substance to another.

  9. Balance a chemical equation • Begin with atoms that appear in only one compound on the left and right. • If an atom occurs as a free element, balance it last. • Change only coefficients (not formulas). C3H8(g) + O2(g)  CO2(g) + H2O(g) last

  10. Formula Weight of NaCl: 23 amu Na + 35.5 amu Cl = 58.5 amu NaCl Molecular Weight of H2O: 2 (1 amu H) + 16 amu O = 18 amu H2O Formula and Molecule Ionic & covalent compounds Formula formula of NaCl Covalent compounds Molecule molecule of H2O

  11. Mole Mole (mol):formula weight of a substance (in gram). 12g of C = 1 mol C 23g of Na = 1 mol Na 58.5 g of NaCl = 1 mol NaCl 18 g of H2O = 1 mol of H2O

  12. Molar mass (g/mol): mass of 1 mole of substance (in gram) (Formula weight) molar mass of Na = 23 g/mol molar mass of H2O = 18 g/mol Avogadro’s number (6.02×1023):number of formula units in one mole. 1 mole of apples = 6.02×1023 apples 1 mole of A atoms = 6.02×1023 atoms of A 1 mole of A molecules = 6.02×1023 molecules of A 1 mole of A ions = 6.02×1023 ions of A

  13. 2 grams 2 grams 1 gram Stoichiometry Relationships between amounts of substances in a chemical reaction. Look at the Coefficients! 2H2O(l) 2H2(g) + O2(g) 2 2 1 2 moles 2 moles 1 mole 2 liters 2 liters 1 liter 2 particles 1 particle 2 particles

  14. volume volume B mole A mole mass mass Particle (atom) (molecule) (ion) Particle (atom) (molecule) (ion) 1 step: use coefficient in the balanced equation. CH4 + 2O2 CO2 + 2H2O

  15. 1 cc CO2 10 cc O2 = ? cc CO2 ) = 5 cc CO2 10 cc O2 ( 2 cc O2 1 mole CH4 1 mole CO2 32 g CH4 = ? moles CO2 )( ) = 2 mole CO2 32 g CH4 ( 16 g CH4 1 mole CH4 1 mole CH4 22.4 L CH4 40 g CH4 = ? L CH4 )( ) = 56 L CH4 40 g CH4 ( 16 g CH4 1 mole CH4 STP: 1 mole of substance (gas) = 22.4 L = 22400 cc (cm3 or mL) CH4 + 2O2 CO2 + 2H2O 2 moles H2O 23 mole CH4 = ? moles H2O 23 mole CH4 ( ) = 46 moles H2O 1 mole CH4

  16. Limiting Reagents N2(g) + O2(g)  2NO(g) 1 mole Stoichiometry: 1 mole 2 moles Before reaction: 1 mole 4 moles After reaction: 0 mole 3 moles 2 moles

  17. Left over Limiting Reagents N2(g) + O2(g)  2NO(g) 1 mole Stoichiometry: 1 mole 2 moles Before reaction: 1 mole 4 moles After reaction: 0 mole 3 moles 2 moles Used up first

  18. Left over Limiting Reagents Limiting reagent N2(g) + O2(g)  2NO(g) 1 mole Stoichiometry: 1 mole 2 moles Before reaction: 1 mole 4 moles After reaction: 0 mole 3 moles 2 moles Used up first

  19. Limiting Reagents Limiting reagent: is the reactant that is used up first. Limiting reagents can control a reaction: N2(g) + O2(g)  2NO(g)

  20. 12g of C 64g of O2 ? Limiting reagent ? g of CO2 will be formed C(s) + O2(g)  CO2(g) C(s) + O2(g)  CO2(g) 1 mol 1 mol 1 mol 1 mole C 12g C( ) = 1 mole C 12g C C is the limiting reagent. 1 mole O2 64g O2 ( ) = 2 mole O2 32g O2 Limiting Reagents Example: Make sure that the chemical equation is balanced.

  21. 1 mole CO2 1 mole C 44g CO2 12g C( )( )( ) = 44g CO2 12g C 1 mole C 1 mole CO2 Limiting Reagents A B ? g of CO2 will be formed: C(s) + O2(g)  CO2(g) 12g 64g ? g We should use the mass of the limiting reagent. (because it controls our reaction).

  22. actual yield × 100 Percent yield = theoretical yield N2(g) + O2(g)  2NO(g) 37g NO theoretical yield = 40g NO actual yield = 37g NO Percent yield = × 100 = 92.5% 40g NO 7.5% error or lost Percent Yield actual yield: mass of product formed (experimental) theoretical yield: mass of product that should form (according to stoichiometry)

  23. H2O AgNO3(s) Ag+(aq) + NO3-(aq) AgNO3 NaCl H2O NaCl(aq) + AgNO3(aq)  AgCl(s) + NaNO3(aq) Aqueous Solution(ionic compounds) H2O NaCl(s) Na+(aq) + Cl-(aq) Dissociation (Ionization) aqueous solution: solvent is water

  24. total charge on left side = total charge on right side 2As3+(aq) + 3s2-(aq)  As2S3(s) Molecular equation: NaCl(aq) + AgNO3(aq)  AgCl(s) + NaNO3(aq) Ionic equation: Na+(aq) + Ag+(aq) + Cl-(aq) + NO3-(aq)  AgCl(s) + Na+(aq) + NO3-(aq) Na+(aq) + Ag+(aq) + Cl-(aq) + NO3-(aq)  AgCl(s) + Na+(aq) + NO3-(aq) Spectator ions Ag+(aq) + Cl-(aq)  AgCl(s) Net ionic equation:

  25. Oxidation and Reduction reactions (redox) oxidation:is the loss of electrons. reduction:is the gain of electrons. Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) redox reaction Zn(s)  Zn2+(aq) + 2e- Zn is oxidized (reducing agent) Cu2+(aq) + 2e-  Cu(s) Cu2+ is reduced (oxidizing agent)

  26. C gains O and loses H is oxidized (reducing agent) O gains H Is reduced (oxidizing agent) Oxidation and Reduction reactions (redox) oxidation:is the gain of oxygen / loss of hydrogen. reduction:is the loss of oxygen / gain of hydrogen. CH4(s) + 2O2(g)  CO2(g) + 2H2O(g) redox reaction single replacement reactions and combustion reactions redox reactions double replacement reactions  non redox

  27. Heat of reaction 2HgO(s) + heat (energy)  2Hg(l) + O2(g) Endothermic reaction C3H8(s) + 5O2(g)  3CO2 + 4H2O + heat (energy) Exothermic reaction All combustion reactions are exothermic.