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Types of Chemical Reactions and Solution Stoichiometry

Types of Chemical Reactions and Solution Stoichiometry. Solutions. two parts solute substance dissolved solvent substance doing the dissolving electrical conductivity electrolyte conducts electricity strong electrolyte many ions available weak electrolyte few ions available

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Types of Chemical Reactions and Solution Stoichiometry

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  1. Types of Chemical Reactions and Solution Stoichiometry

  2. Solutions two parts • solute substance dissolved • solvent substance doing the dissolving electrical conductivity • electrolyte conducts electricity • strong electrolyte many ions available • weak electrolyte few ions available • nonelectrolyte does not allow current to flow

  3. Water as a solvent dissolves many substances because of its chemistry polar covalent bond / polar molecule Hydration--process in which a salt “falls apart” • partial positive (d+) charge on hydrogen is attracted to negative anions • partial negative (d-) charge on oxygen is attracted to positive cations

  4. How does water dissolve solutes?

  5. Strong acids / strong bases HCl HBr Group I cations HI H2SO4 Group II except Mg HNO3 HClO4 HClO3 Completely dissociate into ions 100 molecules of HCl  100 H+ and 100 Cl- NaOH(aq) Na+ + OH-

  6. Strong Acids and Bases

  7. Weak acids /weak bases Everything that was not on the former slide. Small degree of dissociation (few ions) HC2H3O2(aq)H+(aq) + C2H3O2-(aq) NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

  8. Weak Acids/ Weak Bases

  9. Concentration of solutions Molarity = M = moles of solute liter of solution Calculate the molarity of a solution prepared by dissolving 1.56 g of HCl in water to make 26.8 mL of solution. 1.56 g HCl 1 mol HCl = 0.0427 mol HCl 36.5 g HCl 26.8 mL  0.0268 L 0.0427 mol = 1.59 M 0.0268 L

  10. Concentrations of Ions Must decide how the substance dissociates 0.50 M Co(NO3)2 Co(NO3)2 Co2+ + 2NO3- 0.50 M Co2+ and (2 X 0.50) = 1.0 M NO3- Very important relationship !!!!! Liters of solution X mol = moles solute liter

  11. Soluble Salts

  12. Which of the following solutions contains the largest number of ions? 100 mL of 0.100 M NaOH 50.0 mL of 0.200 M BaCl2 75.0 mL of 0.150 M Na3PO4 0.100L X 0.100 mol NaOH = .010 mol Na+ and .010 mol OH- L .050 L X .200 mol BaCl2 = 0.010 mol Ba2+ and 0.020 mol Cl- L 0.075 L X 0.150 mol Na3PO4 = 0.034 mol Na+ and 0.011 mol PO43- L

  13. Dilutions M1V1 = M2V2 How do you prepare 2.50 L of 0.50 M solution of NiCl2 from a 1.00 M stock solution? 2.50 L X 0.50 M = ? X 1.00 M ? = 1.25 L

  14. Precipitation Reactions Know your solubility rules! molecular equationoverall stoichiometry but not necessarily the actual forms in solution complete ionic equationrepresents all ions that are strong electrolytes net ionic equation includes only the changing ions AgNO3 + MgCl2 AgCl + MgNO3 Ag+ + NO3- + Mg2+ + 2Cl-  AgCl + Mg2+ + NO3- Ag+ + 2Cl-  AgCl

  15. Stoichiometry for reactions in solution • Determine what reaction occurs • Write the balanced net ionic equation • Determine which reactant is limiting • Calculate moles or grams of products as needed

  16. Acid-Base Reactions Decide if acid/base is strong or weak Strong acid + strong base HCl(aq) + NaOH(aq) NaCl(aq) + HOH(l) H+ + Cl- + Na+ + OH-  Na+ + Cl- + HOH(l) H+ + OH- H2O Strong acid + weak base 2HCl(aq) + Mg(OH)2(s)  MgCl2(aq) + 2HOH(l) 2H+(aq)+ 2Cl-(aq) + Mg(OH)2(s)  Mg2+(aq) + 2Cl-(aq) + 2HOH(l) 2H+(aq) + Mg(OH)2(s) Mg2+(aq) + 2HOH(l)

  17. Oxidation and Reduction ReactionsReactions in which electrons are transferred Rules for assigning oxidation numbers • Oxidation state of an atom in an element is 0. • Oxidation state of monatomic ion is the same as its charge. • In compounds, fluorine is always -1. • Oxygen is usually -2. exceptions: peroxides (O22-) ox # = -1 and OF2 ox # = +2 • In covalent compounds, hydrogen is +1. • Sum of oxidation numbers must equal the charge of the ion.

  18. Assign oxidation numbers to each atom. Na2CO3 Na = +1 2 X +1 = +2 total C = ? O = -2 3 X -2 = -6 total  C = +4 because +2 +4 -6 0

  19. Oxidation and reduction occur together. One substance must be oxidized while another must be reduced. 2Na(s) + Cl2(g) 2NaCl(s) Na changes from 0 to +1  loses electrons Cl changes from 0 to -1  gains electrons oxidation is the loss of electrons OIL reduction is the gain of electrons RIG

  20. Substance oxidized is the reducing agent. Substance reduced is the oxidizing agent. 2Na(s) + Cl2(g) 2NaCl(s) Na changes from 0 to +1  oxidized Na is reducing agent Cl changes from 0 to -1  reduced Cl2 is oxidizing agent

  21. Half reactions 2PbS(s) + 3O2(g) 2PbO(s) + 2SO2(g) Assign ox. # and determine which is oxidized and which is reduced. S-2  S+4 + 6e-  ox O20 + 2e-  O-2  red PbS is the reducing agent. O2 is the oxidizing agent. Notice that the whole substance is written as the agent.

  22. Balancing Redox Reactions in Acidic Solutions Cr2O72- + Cl- Cr3+ + Cl2 • Pull out half reactions and balance ions Cr2O72- 2Cr3+ 2Cl-  Cl20 • Add H2O and H+ to balance oxygen and hydrogen 14H+ + Cr2O72- 2Cr3+ + 7H2O • Add e- to balance charge (e- lost must = e- gained) 6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O x32Cl-  Cl20 + 2e- • Combine and reduce 6Cl- + 14H+ + Cr2O72-  2Cr3+ + 3Cl2 + 7H2O 6 3 6

  23. Balancing redox reactions in basic solution MnO4- + S2- MnS + S • Follow same steps as for acidic solutions x2 7e- + 8H+ + MnO4- MnS + 4H2O x7 S2-  S0 + 2e- 16H+ + 2MnO4- + 7S2-  2MnS + 8H2O + 7S • Add OH- to both sides to neutralize the H+ ; combine and reduce 16OH- + 16H+ + 2MnO4- + 7S2-  2MnS + 8H2O + 7S + 16OH- 8H2O + 2MnO4- + 7S2-  2MnS +7S + 16OH- 14 16 2 2 8 7 7 14

  24. Electrochemical CellsGalvanic or Voltaic Cell Separating the oxidizing agent from the reducing agent allows the reaction to change chemical energy to electrical energy.

  25. Oxidation – at anode [an ox] Reduction – at cathode [red cat] Electrons flow from cathode to anode [fat cat] Salt bridge – allows e- to flow without mixing solutions

  26. Cell potentialdriving force or pull on e- Use standard reduction potential table to determine potentials Most positive gets reduced Al + Zn2+ Zn + Al3+ From table Al3+ + 3e- Al -1.66 V Zn2+ + 2e-  Zn-0.76 V • Switch the Al half reaction, balance and add Al  Al3+ + 3e- 1.66 V Zn2+ + 2e-  Zn-0.76 V 2Al + 3Zn2+  2Al3+ + 3Zn 0.90 V

  27. http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/electroChem/voltaicCell20.htmlhttp://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/electroChem/voltaicCell20.html

  28. Examples: Fe3+ + Cu  Cu2+ + Fe2+ Ag+ + Cd2+  Ag + Cd Cr3+ + Cl2  Cr2O72- + Cl- [Cr2O72- + 6e-  Cr3+ 1.33 V]

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