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Stoichiometry. Section 11.1 Defining Stoichiometry Section 11.2 Stoichiometric Calculations Section 11.3 Limiting Reactants Section 11.4 Percent Yield. Click a hyperlink or folder tab to view the corresponding slides. Exit. Chapter Menu. Section 11.1 Defining Stoichiometry.

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  1. Stoichiometry Section 11.1Defining Stoichiometry Section 11.2Stoichiometric Calculations Section 11.3Limiting Reactants Section 11.4Percent Yield Click a hyperlink or folder tab to view the corresponding slides. Exit Chapter Menu

  2. Section 11.1 Defining Stoichiometry Describethe types of relationships indicated by a balanced chemical equation. reactant:the starting substance in a chemical reaction Statethe mole ratios from a balanced chemical equation. stoichiometry mole ratio The amount of each reactant present at the start of a chemical reaction determines how much product can form. Section 11-1

  3. Particle and Mole Relationships Chemical reactions stop when one of the reactants is used up. (or the reaction reaches equilibrium) Stoichiometryis the study of quantitative relationships between the amounts of reactants used and amounts of products formed by a chemical reaction. Section 11-1

  4. Particle and Mole Relationships (cont.) Stoichiometry is based on the law of conservation of mass. The mass of reactants equals the mass of the products. Section 11-1

  5. Particle and Mole Relationships (cont.) Section 11-1

  6. animation

  7. Do question 1 page 356 orally

  8. Particle and Mole Relationships (cont.) A mole ratio is a ratio between the numbers of moles of any two substances in a balanced equation. The number of mole ratios that can be written for any equation is (n)(n – 1) where n is the number of species in the chemical reaction. In stoichiometry use the one that is needed for the equation Section 11-1

  9. Do question 3 page 357 orally Do question 7a and b page 357 using molecular model kits

  10. A B C D Section 11.1 Assessment Which of the following is a correct mole ratio for the following equation? 2Al(s) + 3Br2(l) → 2AlBr3(s) A.2 mol Al : 3 mol Br B.3 mol Br2 : 2 mol Al C.2 mol AlBr3 : 1 mol Br2 D.2 mol Br : 2 mol Al Section 11-1

  11. A B C D Section 11.1 Assessment How many mole ratios can be written for the following reaction? 4H2(g) + O2(g) → 2H2O(l) A.6 B.4 C.3 D.2 Section 11-1

  12. Section 1 quiz

  13. Section 11.1 Defining Stoichiometry Key Concepts Balanced chemical equations can be interpreted in terms of moles, mass, and representative particles (atoms, molecules, formula units). The law of conservation of mass applies to all chemical reactions. Mole ratios are derived from the coefficients of a balanced chemical equation. Each mole ratio relates the number of moles of one reactant or product to the number of moles of another reactant or product in the chemical reaction. Study Guide 1

  14. End of Section 11-1

  15. Section 11.2 Stoichiometric Calculations Listthe sequence of steps used in solving stoichiometric problems. chemical reaction:a process in which the atoms of one or more substances are rearranged to form different substances Solvestoichiometric problems. The solution to every stoichiometric problem requires a balanced chemical equation. Section 11-2

  16. Using Stoichiometry • All stoichiometric calculations begins with a balanced chemical equation. 4Fe(s) + 3O2(g)  2Fe2O3(s) Section 11-2

  17. Using Stoichiometry (cont.) Steps to solve mole-to-mole, mole-to-mass, and mass-to-mass stoichiometric problems Complete Step 1 by writing the balanced chemical equation for the reaction. To determine where to start your calculations, note the unit of the given substance. If mass (in grams) of the given substance is the starting unit, begin your calculations with Step 2. If amount (in moles) of the given substance is the starting unit, skip Step 2 and begin your calculations with Step 3. Section 11-2

  18. Using Stoichiometry (cont.) 3.The end point of the calculation depends on the desired unit of the unknown substance. If the answer must be in moles, stop after completing Step 3. If the answer must be in grams, stop after completing Step 4. Section 11-2

  19. Using Stoichiometry (cont.) Section 11-2

  20. Moles to moles Do questions 9+10 page 359 Answers page 937

  21. Moles to mass or vice versa Do questions 11 and 12 page 360 Answers page 937

  22. Mass to mass Do questions 13 and 14 page 362 Answers page 937

  23. A B C D Section 11.2 Assessment A chemical reaction equation must be ____ in order to make stoichiometric calculations. A.measured B.controlled C.balanced D.produced Section 11-2

  24. A B C D Section 11.2 Assessment How many moles of CO2 will be produced in the following reaction if the initial amount of reactants was 0.50 moles? 2NaHCO3 → Na2CO + CO2 + H2O A.0.25 B.0.3 C.0.5 D.1.0 Section 11-2

  25. Section 2 quiz

  26. Section 11.2 Stoichiometric Calculations Key Concepts Chemists use stoichiometric calculations to predict the amounts of reactants used and products formed in specific reactions. The first step in solving stoichiometric problems is writing the balanced chemical equation. Mole ratios derived from the balanced chemical equation are used in stoichiometric calculations. Stoichiometric problems make use of mole ratios to convert between mass and moles. Study Guide 2

  27. End of Section 11-2

  28. Section 11.3 Limiting Reactants Identifythe limiting reactant in a chemical equation. Identifythe excess reactant, and calculate the amount remaining after the reaction is complete. Calculatethe mass of a product when the amounts of more than one reactant are given. molar mass:the mass in grams of one mole of any pure substance Section 11-3

  29. Section 11.3 Limiting Reactants (cont.) limiting reactant excess reactant A chemical reaction stops when one of the reactants is used up. Section 11-3

  30. Why do reactions stop? Reactions proceed until one of the reactants is used up and one is left in excess. The limiting reactant limits the extent of the reaction and, thereby, determines the amount of product formed. The excess reactantsare all the leftover unused reactants. Section 11-3

  31. Why do reactions stop? (cont.) Determining the limiting reactant is important because the amount of the product formed depends on this reactant. Section 11-3

  32. animation

  33. Calculating the Product when a Reactant is Limiting S8(l) + 4Cl2(g) → 4S2Cl4(l) 200.0g S and 100.0g Cl2 Determine which is the limiting reactant • mole of reactants: 1.1410 mol CI2; 0.7797 mol S8 • mole ratios determine that for every 1 mol of S8, 1.808 mol CI2 are available • mole ratio from equation is 4 mol Cl2 : 1 mol S8 Chlorine is limiting since there are less moles available than required by the equation. Section 11-3

  34. Calculating the Product when a Reactant is Limiting (cont.) Calculating the amount of product formed • Multiply the amount of limiting reactant (Cl2) by the mole ratio relating S2Cl2 to Cl2. • 190.4g S2Cl2 form Section 11-3

  35. An alternate way is to calculate the amount of product formed from each amount of reactant and pick the smallest Do questions 20-21 page 368 Answers on page 937

  36. Calculating the Product when a Reactant is Limiting (cont.) Analyzing the excess reactant • Moles reacted • Multiply the moles of Cl2 used by the mole ratio relating S8 to Cl2. • 0.3525 mol S8 • Mass reacted. • Multiply moles reacted by molar mass. • 90.42g of S8 • Excess remaining. • 200.0g – 90.42g = 109.6 g S8 in excess Section 11-3

  37. Calculating the Product when a Reactant is Limiting (cont.) Using an excess reactant can speed up the reaction. Using an excess reactant can drive a reaction to completion. Section 11-3

  38. A B C D Section 11.3 Assessment The mass of the final product in a chemical reaction is based on what? A.the amount of excess reactant B.the amount of limiting reactant C.the presence of a catalyst D.the amount of O2 present Section 11-3

  39. A B C D Section 11.3 Assessment What is the excess reactant in the following reaction if you start with 50.0g of each reactant? P4(s) + 5O2(g) → P4O10(s) A.O2 B.P4 C.Both are equal. D.unable to determine Section 11-3

  40. Section 3 quiz

  41. Section 11.3 Limiting Reactants Key Concepts The limiting reactant is the reactant that is completely consumed during a chemical reaction. Reactants that remain after the reaction stops are called excess reactants. To determine the limiting reactant, the actual mole ratio of the available reactants must be compared with the ratio of the reactants obtained from the coefficients in the balanced chemical equation. Stoichiometric calculations must be based on the limiting reactant. Study Guide 3

  42. End of Section 11-3

  43. Section 11.4 Percent Yield Calculatethe theoretical yield of a chemical reaction from data. process:a series of actions or operations Determinethe percent yield for a chemical reaction. theoretical yield actual yield percent yield Percent yield is a measure of the efficiency of a chemical reaction. Section 11-4

  44. How much product? Laboratory reactions do not always produce the calculated amount of products. Reactants stick to containers. Competing reactions form other products. Section 11-4

  45. How much product? (cont.) The theoretical yieldis the maximum amount of product that can be produced from a given amount of reactant. The actual yield is the amount of product actually produced when the chemical reaction is carried out in an experiment. The percent yieldof a product is the ratio of the actual yield expressed as a percent. Section 11-4

  46. Percent Yield in the Marketplace Percent yield is important in the cost effectiveness of many industrial manufacturing processes. Do 27-28 page 372 answer on page 938 Section 11-4

  47. A B C D Section 11.4 Assessment The amount of product that can be produced from a given amount of reactants based on stoichiometric calculations is: A.actual yield B.percent yield C.theoretical yield D.stoichiometric yield Section 11-4

  48. A B C D Section 11.4 Assessment You calculate the theoretical yield of a chemical reaction starting with 50.0g of reactant is 25.0g of product. What is the percent yield if the actual yield is 22.0g of product? A.88% B.44% C.50% D.97% Section 11-4

  49. Section 4 quiz

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