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Ch 4. Chemical Quantities and Aqueous Reactions

Ch 4. Chemical Quantities and Aqueous Reactions. 1 mol. 2 mol. 1 mol. 2 mol. CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O (g). Stoichiometry of the reaction. FIXED ratio for each reaction. If the amount of one chemical is known we can calculate

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Ch 4. Chemical Quantities and Aqueous Reactions

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  1. Ch 4. Chemical Quantities and Aqueous Reactions

  2. 1 mol 2 mol 1 mol 2 mol CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g) Stoichiometry of the reaction FIXEDratio for each reaction If the amount of one chemical is known we can calculate how much other chemicals are required or produced.

  3. CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g) 1 mol 2 mol 1 mol 2 mol 2 mol 4 mol 2 mol 4 mol 3 mol 6 mol 3 mol 6 mol 5.22 mol 10.44 mol 5.22 mol 10.44 mol

  4. 4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g) 6.02 mol of NH3 is used in the above reaction. How many moles of O2 is required to react with all the NH3? How many moles of H2O will be produced?

  5. 2NH3(g) + 3CuO(s) → N2(g) + 3Cu(s) + 3H2O(g) 2 mol 3 mol 1 mol 3 mol 3 mol 6.02 mol x y z u

  6. 2NH3(g) + 3CuO(s) → N2(g) + 3Cu(s) + 3H2O(g) 2 mol 3 mol 1 mol 3 mol 3 mol 6.04 g mass? mass? x y 6.04 g ÷ (14.01 g/mol x 1 + 1.008 g/mol x 3) = 0.355 mol 0.355 mol Mass of CuO = 0.533 mol x 79.55 g/mol = 42.4 g Mass of N2 = 0.178 mol x 28.02 g/mol = 4.99 g

  7. Other Examples: page 143 ― 144

  8. 1 mol 1 mol 0 mol initial: 0 mol ? mol ? mol ? mol final: ? mol CH4 + 2O2 → CO2 + 2H2O 1 mol 2 mol 0 mol initial: 0 mol 0 mol 0 mol 1 mol final: 2 mol The actual amount of reactants consumed and actual amount of products produced agree with the stoichiometry.

  9. 1 mol 1 mol 0 mol 0 mol initial: CH4 + 2O2 → CO2 + 2H2O x = 0.5 mol 1 mol consumed: 0 mol y z (1 − 0.5) mol = 0.5 mol = 1 mol final: x= 0.5 mol 1 mol CH4 requires 2 mol O2, available O2 is 1 mol: limiting reagent. Result: 1 mol O2 will be consumed completely. CH4 will have leftover: excess reagent.

  10. The reactant of which there are fewer moles than the stoichiometry requires is the limiting reagent. The reactant of which there are more moles than the stoichiometry requires is the excess reagent. Chemical reactions always occur according to the stoichiometry, therefore the limiting reagent is consumed and the excess reagent has leftover. The amount of products is determined by the amounts of reagents that are actually consumed.

  11. 0.5 mol 1 mol consumed: 0 mol (1 − 0.5) mol 0.5 mol 1 mol final: 1 mol 1 mol 0 mol 0 mol initial: = 1 : 2 : 1 : 2 excess reagent limiting reagent CH4 + 2O2 → CO2 + 2H2O 0.5 : 1 : 0.5 : 1

  12. + + + + + + + +

  13. 2 slices of bread + 1 slice if ham → 1 sandwich 4 slices of bread + 1 slice if ham → 1 sandwich + 2 slices of bread excess reagent limiting reagent amount of product excess reagent leftover

  14. For the following reaction, if a sample containing 18.1 g of NH3 reacted with 90.4 g of CuO, which is the limiting reagent? How many grams of N2 will be formed? NH3(g) + CuO(s) → N2(g) + Cu(s) + H2O(g) How many grams of excess reagent will be leftover? If 6.63 g of N2 is actually produced, what is the percent yield?

  15. Procedure for limiting/excess reagent calculations aA + bB → cC + dD • Make sure the equation is balanced. • Find the moles of each reactant: moles = mass in gram / molar mass 3) Pick up any reactant, say A, and use the stoichiometry to calculate the required amount of the other reactant B. • Compare the required amount of B with the available amount of B. a) If required > available, then B is the limiting reagent and A is the excess reagent. b) If required < available, then B is the excess reagent and A is the limiting reagent. • Use the amount of the limiting reagent and the stoichiometry to calculate the amount of any product and the amount of the excess reagent that has been consumed. • Leftover excess reagent = available − consumed • If actual yield is given percent yield = (actually yield / theoretical yield) x 100%

  16. Problem Set 7

  17. Review experiment 11 Review problem sets 6 & 7

  18. Midterm Exam Time: next week during lab session Material covered: up to this point Problem sets are very helpful

  19. Classification of Matter Homogeneous (visibly indistinguishable) (Solutions) Mixtures (multiple components) Heterogeneous (visibly distinguishable) Matter Elements Pure Substances (one component) Compounds

  20. Solute + Solvent = Solution Solvent = water, aqueous solution Water can dissolve many substances

  21. H2O O H H

  22. Solution conducts electricity well

  23. C12H22O11

  24. Solution does not conduct electricity

  25. Solution conducts electricity, but weakly

  26. Based on the electrical conductivity in aqueous solution strong electrolytes electrolytes solutes weak electrolytes nonelectrolytes

  27. strong electrolytes: dissociate 100 % into ions weak electrolytes: only a small fraction dissociate into ions nonelectrolytes: no dissociation

  28. salts: NaCl, K2SO4, …… strong acids: HCl, HNO3, H2SO4, HClO4 strong bases: NaOH, KOH Strong electrolytes Bases: compounds that give OH− when dissolved in water. weak acids: acetic acid: HC2H3O2 weak bases: ammonia: NH3 Weak electrolytes

  29. Reaction of NH3 in Water

  30. concentrations no unit

  31. 10. g of sugar is dissolved in 40. g of water. What is the mass percent of sugar in this solution?

  32. Unit: mol/L or M

  33. Example 4.5, page 153 25.5 g of KBr is dissolved in water and forms a solution of 1.75 L. What is the molarity of the solution? Example 4.6, page 154 How many liters of a 0.125 mol/L NaOH solution contains 0.255 mol of NaOH?

  34. How to prepare 1.00 L of NaCl aqueous solution with a molarity of 1.00 mol/L? 1.00 mol NaCl + 1.00 L of H2O = 1.00 mol/L NaCl (aq)

  35. Solution Dilution Concentrated solutions for storage, called stock solutions stock solution + water  desired solution

  36. moles of solute before dilution = moles of solute after dilution M1V1 = M2V2 M1: molarity of concentrated solution V1: volume of concentrated solution M2: molarity of diluted solution V2: volume of diluted solution Example on page 155 A lab procedure calls for 3.00 L of a 0.500 mol/L CaCl2 solution. How should we prepare it from a 10.0 mol/L stock solution?

  37. Example 4.7, page 156 To what volume should you dilute 0.200 L of a 15.0 mol/L NaOH solution to obtain a 3.00 mol/L NaOH solution?

  38. Types of reactions Precipitation reactions

  39. NaCl(aq) + AgNO3(aq)  AgCl(s) + NaNO3(aq) formula equation Na+(aq) + Cl−(aq) + Ag+(aq) + NO3−(aq)  AgCl(s) + Na+(aq) + NO3−(aq) complete ionic equation Na+(aq) + Cl−(aq) + Ag+(aq) + NO3−(aq)  AgCl(s) + Na+(aq) + NO3−(aq) spectator ions Cl−(aq) + Ag+(aq)  AgCl(s) net ionic equation

  40. EXAMPLE 4.9Predicting whether an Ionic Compound Is Soluble Predict whether each compound is soluble or insoluble. (a) PbCl2 (b) CuCl2 (c)Ca(NO3)2 (d) BaSO4

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