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Chapter 11 – Stoichiometry 11.1What is Stoichiometry? 11.2 Stoichiometric Calculations 11.3Limiting Reactants 11.4Percent Yield.

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Presentation on theme: "Chapter 11 – Stoichiometry 11.1What is Stoichiometry? 11.2 Stoichiometric Calculations 11.3Limiting Reactants 11.4Percent Yield."— Presentation transcript:

1 Chapter 11 – Stoichiometry 11.1What is Stoichiometry? 11.2 Stoichiometric Calculations 11.3Limiting Reactants 11.4Percent Yield

2 Section 11.1 Defining Stoichiometry Describe the types of relationships indicated by a balanced chemical equation. Explain the relationship between stoichiometry and the principle of the conservation of mass. State the mole ratios from a balanced chemical equation. The amount of each reactant present at the start of a chemical reaction determines how much product can form.

3 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. Section 11.1 Defining Stoichiometry

4 Stoichiometry Study of quantitative relationships between amounts of reactants used and amounts of products formed by a chemical reaction Combo of 2 Greek words – “to measure the elements” Really a tool (system) for accounting for the conservation of matter What goes in must come out

5 Rests on principle of conservation of matter (mass) 2 Al(s) + 3 Br 2 (l)  Al 2 Br 6 (s) Stoichiometry

6 Stoichiometry - Questions If I have this much reactant, how much product can I form? If I have this much of one reactant, how much of a second reactant do I need so that after the reaction is completed I won’t have any of either reactant left over?

7 Mass reactant Stoichiometric factor Moles reactant Moles product Mass product Requires balanced equation! Conversion Map for Stoichiometry Calculations

8 Stoichiometry & Balanced Equation 4Fe(s) + 3O 2 (g)  2Fe 2 O 3 (s) Can interpret as 4 atoms Fe + 3 molecules O 2  2 formula units Fe 2 O 3 4 mol Fe + 3 mol O 2  2 mol Fe 2 O 3 No direct information on masses Can use molar mass to get mass info

9 Stoichiometry & Balanced Equation 4Fe(s) + 3O 2 (g)  2Fe 2 O 3 (s) 4 mol Fe  g Fe/mol Fe = g Fe 3 mol O 2  g O 2 /mol O 2 = g O 2 2 mol Fe 2 O 3  g Fe 2 O 3 /mol Fe 2 O 3 = g Fe 2 O 3 Total mass reactants = g Total mass products = g

10 Stoichiometry & Balanced Equation Example problem 11.1, p. 370 C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(g) Interpret as molecules, moles & mass 1 molecule C 3 H molecules O 2  3 molecules CO molecules H 2 O 1 mol C 3 H mol O 2  3 mol CO mol H 2 O 1 mol C 3 H 8  g C 3 H 8 /mol C 3 H 8 = g C 3 H 8 5 mol O 2  g O 2 /mol O 2 = g O 2 3 mol CO 2  g CO 2 /mol CO 2 = g CO 2 4 mol H 2 O  g H 2 O /mol H 2 O = g H 2 O Mass reactants = = g Mass products = = g

11 Practice Problems 1 (a-c), 2(a-b) page 371 Problems page 392 Problems 1- 3, page 982

12 Mole Ratio Ratio between the numbers of moles of any two substances in a balanced chemical equation 2Al(s) + 3Br 2 (l)  2AlBr 3 (s) Possible to write 3 unique mole ratios: Al:Br 2 (2:3) Al: AlBr 3 (2:2) Br 2 :AlBr 3 (3:2) Get 3 more ratios from inverses of above for a total of 6 for this reaction

13 Mole Ratio In general, for n species (reactant, product) in chemical equation, have n(n-1) mole ratios 3Fe(s) + 4H 2 O(l)  Fe 3 O 4 (s) + 4H 2 (g) 4 species, 4  3 = 12 possible mole ratios (6 + inverse of each) Fe:H 2 O Fe:Fe 3 O 4 Fe:H 2 H 2 O:Fe 3 O 4 H 2 O:H 2 Fe 3 O 4 : H 2

14 Mole Ratio How many mole ratios can be written for the following reaction? 4H 2 (g) + O 2 (g) → 2H 2 O(l) A. 6 B. 4 C. 3 D. 2 ?

15 Practice Problems 3 (a-c), 4 (a-b) page 372 Problems , page 392 Problems 4 - 5, page 982-3

16 Chapter 11 – Stoichiometry 11.1What is Stoichiometry? 11.2 Stoichiometric Calculations 11.3Limiting Reactants 11.4Percent Yield

17 Section 11.2 Stoichiometric Calculations List the sequence of steps used in solving stoichiometric problems. Solve stoichiometric problems (mole-to-mole, mole-to- mass, mass-to-mass and variations including number of particles) using the factor-label method. The solution to every stoichiometric problem requires a balanced chemical equation.

18 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. Section 11.2 Stoichiometric Calculations

19 Stoichiometric Calculations Start with balanced equation Determine required mole ratios Use mole-to-mass and mass-to-mole relations for 3 basic conversions: 1)Mole to mole 2)Mole to mass 3)Mass to mass Once have moles, can get # particles

20 Mass reactant Stoichiometric factor Moles reactant Moles product Mass product Requires balanced equation! Conversion Map for Stoichiometry Calculations

21 nA + mB  xC + yD Mass reactant Moles reactant # part. reactant Mass product Moles product # part. product

22 Mole to Mole Conversions Given number of moles of a reactant (product), how many moles of a product (reactant) must be formed? Use initial moles and mole ratio mol known x [mol unknown/mol known] = mol unknown mole ratio from equation

23 Mole to Mole Conversions 2K(s) + 2H 2 O(l)  2KOH(aq) + H 2 (g) Mol H 2 if start with 4.00x10 -2 mol K? 4.00x10 -2 mol K  [1 mol H 2 /2 mol K] = 2.00x10 -2 mol H 2 mole ratio from equation

24 Mole to Mole Conversions How many moles of CO 2 will be produced in following reaction if initial amount of reactants = 0.50 moles? 2NaHCO 3 (s) → Na 2 CO 3 (s) + CO 2 (g) + H 2 O(g) A.0.25 B. 0.3 C. 0.5 D. 1.0 ?

25 Practice Problems 11,12 page 375 Problems , page 393 Problems 6 - 7, page 983

26 Mole to Mass Conversions Step 1 – Do mole to mole conversion using mole ratio Step 2 – Do mole to mass conversion using molar mass

27 Mass reactant Stoichiometric factor Moles reactant Moles product Mass product Requires balanced equation! Conversion Map for Stoichiometry Calculations

28 Mole to Mass Conversions 2Na(s) + Cl 2 (g)  2NaCl(s) Mass NaCl produced from 1.25 mol Cl 2 ? 1.25 mol Cl 2  [2 mol NaCl/1 mol Cl 2 ] mole ratio from equation = 2.50 mol NaCl 2.50 mol NaCl  [58.44 g NaCl/mol NaCl] = 146 g NaCl molar mass of NaCl

29 Practice Problems 13,14 page 376 Problems , page 393 Problems 8 - 9, page 983

30 Mass to Mass Conversions Step 1 – Do mass to mole conversion using molar mass Step 2 – Do mole to mole conversion using mole ratio Step 3 – Do mole to mass conversion using molar mass

31 Mass reactant Stoichiometric factor Moles reactant Moles product Mass product Requires balanced equation! Conversion Map for Stoichiometry Calculations

32 Mass to Mass Conversions NH 4 NO 3 (s)  N 2 O(g) + 2H 2 O(g) Mass of H 2 O from 25.0 g NH 4 NO 3 (s)? AN= NH 4 NO 3 (s) 25.0 g AN  [1 mol AN/80.04 g AN] 1/molar mass of AN = mol AN mol AN  [2 mol H 2 O /1 mol AN] mole ratio from equation = mol H 2 O

33 Mass to Mass Conversions NH 4 NO 3 (s)  N 2 O(g) + 2H 2 O(g) Mass of H 2 O from 25.0 g NH 4 NO 3 (s)? mol H 2 O  [18.02 g H 2 O /mol H 2 O ] molar mass of H 2 O = 11.2 g H 2 O

34 Practice Problems 15,16 page 377 Problems 60, , pages Problems , page 983

35 Chapter 11 – Stoichiometry 11.1What is Stoichiometry? 11.2 Stoichiometric Calculations 11.3Limiting Reactants 11.4Percent Yield

36 Section 11.3 Limiting Reactants Identify the limiting reactant in a chemical equation. Identify the excess reactant, and calculate the amount remaining after the reaction is complete. Calculate the mass of a product when the amounts of more than one reactant are given. A chemical reaction stops when one of the reactants is used up.

37 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. Section 11.3 Limiting Reactants

38 Limiting Reactants Typical case: reactants not present in exact stoichiometric ratios Mole ratio given by coefficients of reactants in balanced chemical equation N 2 + 3H 2  2NH 3 Stoichiometric ratio of reactants for ammonia synthesis is 1 N 2 : 3 H 2 If ratio  1:3 then one reactant is limiting

39 Limiting and Excess Reactants Ammonia Synthesis N 2 (g) + 3H 2 (g)  2NH 3 (g) Case of 1:1 N 2 :H 2 mole ratio Before ReactionAfter Reaction 3N 2 3H 2 2NH 3 2N 2 H 2 is limiting reactant N 2 is excess reactant

40 Limiting and Excess Reactants Limiting reactant Limits the extent of reaction Determines the amount of product Excess reactant Always have “left over” Often a reactant is deliberately used in excess to:  Drive reaction to completion (equilibrium)  Speed up reaction (kinetics)

41 Limiting and Excess Reactants Limiting and excess air (oxygen) in combustion reaction Air limiting – soot (carbon) Air excess – CO 2 & H 2 O

42 Limiting Reactant - Identifying Step 1 – If masses given, convert to moles Step 2 – Compare actual mole ratio of reactants to mole ratio determined from balanced equation Ratios same - no limiting reactant Ratios different - compare to identify limiting reactant

43 Limiting Reactant S 8 (l) + 4Cl 2 (g)  4S 2 Cl 2 (l) g sulfur reacts with g Cl 2 Part a: g of product produced? Step 1 – Convert masses to moles g S 8  [1 mol S 8 /256.5 g S 8 ] = mol S g Cl 2  [1 mol Cl 2 /70.91 g Cl 2 ] = mol Cl 2 1/molar mass of Cl 2 1/molar mass of S 8

44 Limiting Reactant S 8 (l) + 4Cl 2 (g)  4S 2 Cl 2 (l) g sulfur reacts with g Cl 2 Part a: g product ? Step 2 – Compare actual mole ratio to mole ratio from equation = 1.808:1 Equation: Cl 2 :S 8 = 4:1 Actual: mol Cl 2 : mol S 8 Compare ratios  Cl 2 is limiting

45 Limiting Reactant S 8 (l) + 4Cl 2 (g)  4S 2 Cl 2 (l) g sulfur reacts with g Cl 2 Part a: g product ? Cl 2 limiting reactant, have mol Calculate mol product using mole ratio mol Cl 2  [4 mol S 2 Cl 2 /4 mol Cl 2 ] = mol S 2 Cl mol S 2 Cl 2  [135.0 g S 2 Cl 2 /mol S 2 Cl 2 ] Mole ratio molar mass of S 2 Cl 2 = g S 2 Cl 2

46 Limiting Reactant S 8 (l) + 4Cl 2 (g)  4S 2 Cl 2 (l) g sulfur reacts with g Cl ? Part b: g of sulfur actually reacted? Cl 2 limiting reactant, have mol Calculate mol sulfur using mole ratio mol Cl 2  [1 mol S 8 /4 mol Cl 2 ] = mol S mol S 8  [256.5 g S 8 /mol S 8 ] Mole ratio molar mass of S 8 = g S 8 reacts g excess

47 Limiting Reactant S 8 (l) + 4Cl 2 (g)  4S 2 Cl 2 (l) g sulfur reacts with g Cl ? Part b: g of sulfur actually reacted? Know that all Cl 2 (g) reacted (100.0 g) Previously calculated g product S 2 Cl 2 (l) produced Cons. mass: g = starting mass = final mass = m[S 8 (l)] + m[4S 2 Cl 2 (l)] m[S 8 (l)]= g -m[4S 2 Cl 2 (l)] =109.6 g

48 Limiting Reactant Excess reactant in following reaction if start with 50.0g of each reactant? P 4 (s) + 5 O 2 (g) → P 4 O 10 (s) A.O 2 B.P 4 C. None – quantities are stoichiometric ? 50.0 g P 4  [1 mol P 4 / g P 4 ]= mol P g O 2  [1 mol O 2 /32.00 g O 2 ] = 1.56 mol O 2 Need 5  mol = 2.02 mol O 2  O 2 limiting

49 Practice Problems 23, 24 page 383 Problems , page 394 Problems , page 983

50 Chapter 11 – Stoichiometry 11.1What is Stoichiometry? 11.2 Stoichiometric Calculations 11.3Limiting Reactants 11.4Percent Yield

51 Section 11.4 Percent Yield Calculate the theoretical yield of a chemical reaction from data. Explain the difference between a theoretical yield of a chemical reaction and an actual yield. Provide reasons why the theoretical yield and actual yield can be different from each other. Determine the percent yield for a chemical reaction. Explain why the percent yield can exeed 100% Percent yield is a measure of the efficiency of a chemical reaction.

52 Key Concepts The theoretical yield of a chemical reaction is the maximum amount of product that can be produced from a given amount of reactant. Theoretical yield is calculated from the balanced chemical equation. The actual yield is the amount of product produced. Actual yield must be obtained through experimentation. Percent yield is the ratio of actual yield to theoretical yield expressed as a percent. High percent yield is important in reducing the cost of every product produced through chemical processes. Section 11.4 Percent Yield

53 Percent Yield Actual (real world) reactions don’t produce amount of product expected from masses of reactant and the balanced chemical equation (stoichiometry)

54 Types of Yields Theoretical Yield Maximum amount of product that can be produced from given amounts of reactants Calculated using stoichiometry Actual Yield Amount of product actually obtained when the reaction is carried out

55 Types of Yields Theoretical Yield and Actual Yield can differ because Reaction stops before limiting reactant all used up Product is not all collected – sticks to reaction vessel, filter paper, etc Reactions other than the intended reaction occur, producing other products Impurities in reactants

56 Alternate Reaction Paths Combustion of natural gas (methane) CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g) Know this reaction takes place if have enough O 2 – get nice blue flame Know if limit O 2, get yellow flame and soot (carbon) production Same reactants yield different products

57 Percent Yield Percent yield is a way of expressing the efficiency of a reaction Percent yield = 100  actual yield theoretical yield Theoretical yield computed using same procedures used in previous sections (stoichiometry)

58 % Percent Yield Theoretical yield of chemical reaction = 25 g. % yield if actual yield = 22 g ? A. 88% B. 44% C. 50% D. 97% ? % yield = 100  22 g / 25 g = 88%

59 Percent Yield In commercial production of chemicals, attempt to produce as close to 100% yield as possible Optimum use of materials Minimizes or eliminates need to separate product from unused reactants and from unwanted products

60 Percent Yield for Multi-Stage Attempt to produce as close to 100% yield as possible Critical for multi-stage reaction processes where product of one reaction used as one of the reactants of a subsequent reaction A three-stage reaction with 90% yield from each stage gives only 73% overall yield (0.9 x 0.9 x 0.9 = 0.73)

61 >100% Percent Yield Actual Yield: Amount of product actually obtained when from reaction Obtained from a laboratory measurement – most common is measuring mass on a balance Assumption made is that the only “stuff” you’re measuring is the product of interest

62 >100% Percent Yield Possible to get more “stuff” you think is product than predicted if: Haven’t completely separated intended product from other products in reaction or from leftover reactants and/or solvent (insufficient drying) Product reacts with air or water vapor before it is weighed

63 Practice Problems page 387 Problem 35, page 388 Problems , page 395 Problems , page 983


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