CHAPTER THREE CHEMICAL EQUATIONS & REACTION STOICHIOMETRY

Transcription

1 CHAPTER THREE CHEMICAL EQUATIONS & REACTION STOICHIOMETRY 1

2 Chapter Three Goals 1. Chemical Equations. Calculations Based on Chemical Equations. Percent Yields from Chemical Reactions 4. The Limiting Reactant Concept 5. Concentrations of Solutions 6. Dilution of solutions 7. Using Solutions in Chemical Reactions

3 Chemical Equations A chemical process is represented by a chemical equation 1. Reaction of methane with O : CH 4 + O CO + H O reactants products. reactants on left side of reaction. products on right side of equation 4. relative amounts of each using stoichiometric coefficients

4 4 Chemical Equations

5 Chemical Equations Look at the information an equation provides: Fe O + CO Fe + CO reactants yields products 1 formula unit molecules atoms molecules 1 mole moles moles moles g 84.0 g g 1 g 5

6 Chemical Equations Law of Conservation of Matter There is no detectable change in quantity of matter in an ordinary chemical reaction. Balanced chemical equations must always include the same number of each kind of atom on both sides of the equation. Propane,C H 8, burns in oxygen to give carbon dioxide and water. 6 C H + 5 O CO + 4 H O 8

7 Law of Conservation of Matter NH burns in oxygen to form NO & water NH + 5 O NO + H O or correctly 4 NH + 5 O 4 NO + 6 H O 7

8 Law of Conservation of Matter C 7 H 16 burns in oxygen to form carbon dioxide and water. You do it! C H + 11 O 7CO + 8H O 7 16 Balancing equations is a skill acquired only with lots of practice work many problems 8

9 Chemical Equations Look at the information an equation provides: Fe O + CO Fe + CO reactants yields products 1 formula unit molecules atoms molecules 1 mole moles moles moles g 84.0 g g 1 g 9

10 Calculations Based on Chemical Equations How many CO molecules are required to react with 5 formula units of Fe O? 5 Fe O +? CO Product 1Fe O needs CO 10 5Fe O needs? CO? CO molecules = 5 formula units Fe = 75 molecules of CO O CO molecules 1Fe O formula unit

11 Calculations Based on Chemical Equations How many iron atoms can be produced by the reaction of.50 x 10 5 formula units of iron (III) oxide with excess carbon monoxide? Fe O + excess CO Fe + CO 1Fe O gives Fe.5 X 10 5 Fe O gives? Fe 11? Fe atoms = Fe atoms 1 formula units Fe O 5 formula = 5.00 units 10 5 Fe Fe O atoms

12 Calculations Based on Chemical Equations 1 What mass of CO is required to react with 146 g of iron (III) oxide? Fe O + CO Product MW(Fe O ) needs MW(CO) 146 g needs?g CO? g CO = 146 g Fe O 1 mol Fe O g Fe O 8.0 g CO = 76.8 g CO 1 mol CO mol CO 1 mol Fe O

13 Calculations Based on Chemical Equations What mass of carbon dioxide can be produced by the reaction of mole of iron (III) oxide with excess carbon monoxide? Fe O + excess CO Fe + CO 1mol Fe O gives mol CO mol Fe O gives? mol CO? mol CO =? g CO /MW(g/mol) CO 1 mol CO? g CO = mol FeO 1 mol Fe O = 71. g CO g CO mol CO

14 Calculations Based on Chemical Equations What mass of iron (III) oxide reacted with excess carbon monoxide if the carbon dioxide produced by the reaction had a mass of 8.65 grams? You do it! Fe O + excess CO Fe + CO? g Fe O = 9.57 g Fe O 14

15 Percent Yields from Reactions Theoretical yield is calculated by assuming that the reaction goes to completion. Actual yield is the amount of a specified pure product made in a given reaction. 15 In the laboratory, this is the amount of product that is formed in your beaker, after it is purified and dried. Percent yield indicates how much of the product is obtained from a reaction. % yield = actual yield theoretical yield 100%

16 Percent Yields from Reactions A 10.0 g sample of ethanol, C H 5 OH, was boiled with excess acetic acid, CH COOH, to produce 14.8 g of ethyl acetate, CH COOC H 5. What is the percent yield? CH COOH + C H 5 OH CH COOC H 5 + H O MW MW 10.0 g X (Theoretical Yield) 16

17 Percent Yields from Reactions CH COOH? g CH COOC % yield = Calculate the theoretical yield H 5 + C H = 10.0 g C H. Calculate the percent yield g CHCOOC 19.1g CH COOC 5 OH CH COOC 5 OH = 19.1g CH COOC H H 5 5 H 88.0 g CHCOOCH 46.0 g C H OH H H O 100% = 77.5%

18 Percent Yields from Reactions Salicylic acid reacts with acetic anhydride to form aspirin, acetylsalicylic acid. If the percent yield in this reaction is 78.5%, what mass of salicylic acid is required to produce 150. g aspirin? C 7 H 6 O + C 4 H 6 O C 9 H 8 O 4 + H O salicylic acid acetic anhydride MW = 18 g/mol aspirin MW = 180 g/mol 18 9 billion tablets are consumed by Americans each year

19 Percent Yields from Reactions actual Yield (150 g) 78.5 = x 100 Theoretical Yield (g) C 7 H 6 O + C 4 H 6 O C 9 H 8 O 4 + H O salicylic acid acetic anhydride aspirin MW X 19 Answer: MW (Theoretical Yield) X = g

20 0 Limiting Reactant Concept 1. In a given reaction, there is not enough of one reagent to use up the other reagents completely.. The reagent in short supply LIMITS the quantity of the product that can be formed.. How many bikes can be made from 10 frames and 16 wheels? 1 frame + wheels 1 bike excess limiting

21 1 Limiting Reactant Concept

22 Limiting Reactant Concept When g mercury is reacted with g bromine to form mercuric bromide, which is the limiting reagent? Hg + Br HgBr Thus the limiting reagent is

23 Limiting Reactant Concept What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g of carbon disulfide with 110. g of oxygen? CS + O CO + SO 1 mol mol 1 mol mol 76. g (.0 g) 44.0 g (64.1 g)

24 Limiting Reactant Concept 4?molSO?molSO = 95.6g CS = 110g O CS + O 1molO.0g O 1molCS 76.g CO + molso 1molCS molso molo SO 64.1g SO 1molSO 64.1g SO 1molSO = 161g SO = 147g SO Which is limiting reactant? Limiting reactant is O. What is maximum mass of sulfur dioxide? Maximum mass is 147 g.

25 Limiting Reactant Concept PbO + Cr (SO 4 ) + K SO 4 + H O PbSO 4 + K Cr O 7 + H SO 4 If 5.0 g of each reactant were used in performing the following reaction, which would be the limiting reactant? (a) PbO (b) H O (c) K SO 4 (d) PbSO 4 (e) Cr (SO 4 ) MnO + 4KOH + O + Cl KMnO 4 + KCl + H O If 0.0 g of each reactant were used in performing the following reaction, which would be the limiting reactant? 5 (a) MnO (b) KOH (c) O (d) Cl (e) KMnO 4

26 Concentration of Solutions 6 Definition of Solution: a homogeneous mixture of two or more substances dissolved in another. A solution is composed of two parts: (1) Solute: dissolved substance (or substance in the lesser amount). () Solvent: dissolving substance (or substance in the greater amount). In aqueous solutions, the solvent is water. Example: Solution of NaCl in water, H O: NaCl: solute, H O: solvent

27 Concentration of Solutions 7 Amount of solute Concentration = Mass or Volume of solution Relative terms: Dilute solution: small amount of solute in large amount of solvent. Concentrated solution: large amount of solute in smaller amount of solvent (e.g. the amount of sugar in sweet tea can be defined by its concentration). We will discuss concentration units: 1- Percent by mass (do not confuse with % by mass of element in compound) - Molarity

28 Concentration of Solutions 1- Percent by mass: mass of solute % by mass of solute = 100 % mass of solution mass of solution = mass of solute + mass of solvent % by mass of solute has the symbol % w/w Note: if the question says the solution is aqueous oe does not Specify the solvent, the solvent is water, H O. 8

29 Concentration of Solutions Calculate the mass of potassium nitrate, KNO required to prepare 50.0 g of solution that is 0.0 % KNO by mass. What is the mass of water in the solution? (a) % by mass = g KNO g solution x % = g KNO = g KNO x g 0.0 % x 50.0 g 100 = 50.0 g

30 Concentration of Solutions (b) mass of solution = mass of KNO + mass H O mass H O = mass of solution - mass of KNO mass H O = 50.0 g g mass H O = 00.0 g 0

31 Concentration of Solutions Calculate the mass of 8.00% w/w NaOH solution that contains.0 g of NaOH.? g solution =.0 g g solution NaOH 8.00 g NaOH = 400. g sol' n 1

32 Concentration of Solutions Calculate the mass of NaOH in 00.0 ml of an 8.00% w/w NaOH solution. Density is 1.09 g/ml. You do it!? g 1.09 g sol' n NaOH = 00.0 ml sol' n 1mL sol' n 8.00 g NaOH = 6. g NaOH 100 g sol' n

33 Concentrations of Solutions What volume of 1.0% KOH contains 40.0 g of KOH? The density of the solution is 1.11 g/ml. You do it!? ml solution = 40.0 g KOH g solution 1.0 g KOH 1 ml solution 1.11 g solution = 00. ml solution

34 Concentrations of Solutions - Second common unit of concentration: Molarity 4

35 Concentrations of Solutions - Second common unit of concentration: Molarity molarity M M = = = number of number of moles L mmol ml moles of solute liters of solution 5

36 Concentrations of Solutions 6 Calculate the molarity of a solution that contains 1.5 g of sulfuric acid in 1.75 L of solution.? mol H SO L sol' n You do it! 1.5 g H SO = 1.75 L sol' n = = 4 mol H SO L M H SO 1 mol H 98.1 g H 4 4 SO SO 4 4

37 Concentrations of Solutions Determine the mass of calcium nitrate required to prepare.50 L of M Ca(NO ). You do it! mol Ca(NO )? g Ca(NO ) = 50. L L 164 g Ca(NO ) = 459 g Ca(NO ) 1 mol Ca(NO ) 7

38 Concentrations of Solutions One of the reasons that molarity is commonly used is because: M x L = moles solute and M x ml = mmol solute 8

39 Concentrations of Solutions The specific gravity of concentrated HCl is and it is 6.1% w/w HCl. What is its molarity? 9 specific gravity = 1.185? mol density =1.185 g/ml or 1185g/L HCl/L = 1mol HCl 6.46 g HCl tells us 1185 g solution L solution = M HCl 6.1g HCl 100 g sol'n

40 Dilution of Solutions To dilute a solution, add solvent to a concentrated solution. One method to make tea less sweet. The number of moles of solute (before and after dilution) in the two solutions remains constant. The relationship M 1 V 1 = M V is appropriate for 40 dilutions, but not for chemical reactions.

41 Dilution of Solutions Common method to dilute a solution involves the use of volumetric flask, pipet, and suction bulb. 41

42 Dilution of Solutions Take-Home Calculations (M x V) A = (M x V) B M x V = moles of solute M x V = W/MW (M x V) A = (W/MW) A OR 4 (M x V) A = (W/MW) B W = M x V x MW

43 Dilution of Solutions If 10.0 ml of 1.0 M HCl is added to enough water to give 100. ml of solution, what is the concentration of the solution? M M ml M V 1 = = = = M M V ml 1.0 M 10.0 ml ml 1.0 M

44 Dilution of Solutions What volume of 18.0 M sulfuric acid is required to make.50 L of a.40 M sulfuric acid solution? You do it! 44 M 1 V V V = = = = M M V V M 1.50 L M 0. L or M ml

45 Using Solutions in Chemical Reactions Combine the concepts of molarity and stoichiometry to determine the amounts of reactants and products involved in reactions in solution. 45

46 Using Solutions in Chemical Reactions 46 What volume of M BaCl is required to completely react with 4. g of Na SO 4? Na SO? L BaCl 1mol BaCl 1mol Na SO 4 = + BaCl 4. gna 4 SO BaSO 1mol 14 g 1L BaCl mol BaCl = NaCl Na Na SO SO L

47 Using Solutions in Chemical Reactions (a)what volume of 0.00 M NaOH will react with 50.0 ml 0f 0.00 M aluminum nitrate, Al(NO )? ( NO ) + NaOH Al( OH) Al + NaNO You do it! 47

48 Using Solutions in Chemical Reactions (a) What volume of 0.00 M NaOH will react with 50.0 ml 0f 0.00 M aluminum nitrate? 48? Al ml ( NO ) NaOH = 0.00 mol Al(NO 1 L Al(NO ) 1 L NaOH 0.00 mol NaOH NaOH = L Al(OH) or 150 ml NaNO 1 L ml Al(NO ) sol' n 1000 ml ) sol' n mol NaOH sol' n 1 mol Al(NO ) + NaOH sol' n

49 Using Solutions in Chemical Reactions (b) What mass of Al(OH) precipitates in (a)? 1 L? g Al(OH) = 50.0 ml Al(NO) sol' n 1000 ml 0.00 mol Al(NO) 1mol Al(OH) 78.0 g Al(OH) 1 L Al(NO ) sol' n 1mol Al(NO ) 1mol Al(OH) = g Al(OH) 49

50 Homework Assignment One-line Web Learning (OWL): Chapter Exercises and Tutors Optional 50

51 51 End of Chapter