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CHEMICAL REACTIONS CHAPTER 4 Reactants: Zn + I 2 Product: ZnI 2.

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Presentation on theme: "CHEMICAL REACTIONS CHAPTER 4 Reactants: Zn + I 2 Product: ZnI 2."— Presentation transcript:

1 CHEMICAL REACTIONS CHAPTER 4 Reactants: Zn + I 2 Product: ZnI 2

2 2 Chapter 4 Outline Chemical Equations Stoichiometry –Limiting Reactants –Chemical Analysis

3 3 Chemical Equations Depict the kind of reactants and products and their relative amounts in a reaction 4 Al(s) + 3 O 2 (g) 2Al 2 O 3 ( s) The numbers in the front are called stoichiometric coefficients The letters (s), (g), and (l) are the physical states of compounds.

4 4 Chemical Equations 4 Al(s) + 3 O 2 (g) 2 Al 2 O 3 (s) This equation means 4 Al atoms + 3 O 2 molecules 2 molecules of Al 2 O 3 4 moles of Al + 3 moles of O 2 2 moles of Al 2 O 3

5 5 Chemical Equations Because the same atoms are present in a reaction at the beginning and at the end, the amount of matter in a system does not change. Law of the Conservation of MatterThe Law of the Conservation of Matter

6 6 conservation of matter equation must be balanced. Because of the principle of the conservation of matter, an equation must be balanced. It must have the same number of atoms of the same kind on both sides. Chemical Equations Lavoisier, 1788

7 7 Balancing Equations

8 8

9 9 ___C 3 H 8 (g) + ___ O 2 (g) ----> ___CO 2 (g) + ___ H 2 O (g) C 3 H 8 (g) + 5 O 2 (g) ----> 3 CO 2 (g) + 4 H 2 O (g)

10 10 Balancing Equations ___B 4 H 10 (g) + ___ O 2 (g) ---> ___ B 2 O 3 (g) + ___ H 2 O (g) 2 B 4 H 10 (g) + 11 O 2 (g) ---> 4 B 2 O 3 (g) + 10 H 2 O (g) 2 B 4 H 10 (g) + 11 O 2 (g) ---> 4 B 2 O 3 (g) + 10 H 2 O (g)

11 STOICHIOMETRY the study of the quantitative aspects of chemical reactions. - the study of the quantitative aspects of chemical reactions. 11

12 12 STEP 1 Write the balanced chemical equation NH 4 NO 3 N 2 O + 2 H 2 O PROBLEM: If 454 g of NH 4 NO 3 decomposes, how much H 2 O and N 2 O are formed? What is the theoretical yield of products?

13 g of NH 4 NO 3 --> N 2 O + 2 H 2 O STEP 2 Convert reactant mass to moles (454 g) --> moles 454 g 1 mol g = 5.68 mol NH 4 NO 3

14 g of NH 4 NO 3 --> N 2 O + 2 H 2 O STEP 3 Convert moles reactant --> moles product. Relate moles NH 4 NO 3 to moles product. 1 mol NH 4 NO 3 --> 2 mol H 2 O Express this relation as the STOICHIOMETRIC FACTOR STOICHIOMETRIC FACTOR 2 mol H 2 O produced 1 mol NH 4 NO 3 used

15 g of NH 4 NO 3 --> N 2 O + 2 H 2 O = 11.4 mol H 2 O produced STEP 3 Convert moles reactant (5.68 mol) moles product 5.68 mol NH 4 NO 3 2 mol H mol H2 O produced 1 mol NH 4 NO 3 used used

16 g of NH 4 NO 3 --> N 2 O + 2 H 2 O STEP 4 Convert moles product (11.4 mol) to mass product. This is called the THEORETICAL YIELD This is the “Expected” # of moles.

17 g of NH 4 NO 3 --> N 2 O + 2 H 2 O This is the “Expected” Mass! ALWAYS FOLLOW THESE STEPS IN SOLVING STOICHIOMETRY PROBLEMS! Repeat to find the grams of N 2 O formed. STEP 4 Convert moles prod. (11.4 mol) to mass prod g O 11.4 mol H 2 O 1 mol = 204 g H 2

18 g of NH 4 NO 3 --> N 2 O + 2 H 2 O STEP 5 How much N 2 O is formed? Total mass of reactants=total mass of products 454 g NH 4 NO 3 = ___ g N 2 O g H 2 O mass of N 2 O = 250. g This is an alternate method.

19 g of NH 4 NO 3 --> N 2 O + 2 H 2 O STEP 6 Calculate the percent yield. If you isolated only 131 g of N 2 O, what is the percent yield? This compares the theoretical (250. g) and actual (131 g) yields.

20 g of NH 4 NO 3 --> N 2 O + 2 H 2 O STEP 6 Calculate the percent yield. % yield yield= actual yield theoretical yield 100% 100% % yield yield= 131 g 250. g 100% 100%= 52.4% 52.4%

21 21 Mass reactant Moles reactant Moles product Mass product General Plan For Stoichiometry Calculations

22 22 PROBLEM: Using 5.00 g of H 2 O 2, what mass of O 2 and of H 2 O can be obtained? 2 H 2 O 2 (liq) ---> 2 H 2 O (g) + O 2 (g) Reaction is catalyzed by MnO 2 Step 1: moles of H 2 O 2 Step 2: use STOICHIOMETRIC FACTOR to calculate moles of O 2 Step 3: mass of O 2 Repeat for H 2 O.

23 Reactions Involving a LIMITING REACTANT In a given reaction, there is not enough of one reagent to use up the other reagent completely.In a given reaction, there is not enough of one reagent to use up the other reagent completely. The reagent in short supply LIMITS the quantity of product that can be formed.The reagent in short supply LIMITS the quantity of product that can be formed.

24 24 LIMITING REACTANTS R eactantsProducts 2 NO(g) + O 2 (g) 2 NO 2 (g) Limiting reactant = ___________ Excess reactant = ____________ NONO O2O2

25 25 LIMITING REACTANTS

26 26 Rxn 1 Rxn 2 Rxn 3 Rxn 1 Rxn 2 Rxn 3 mass Zn (g) mol Zn mol HCl mol HCl/mol Zn LIMITING REACTANTS React solid Zn with mol HCl (aq) Zn + 2 HCl ---> ZnCl 2 + H 2

27 27 PROBLEM: Mix 5.40 g of Al with 8.10 g of Cl 2. How many grams of Al 2 Cl 6 can form?

28 28 Step 1 of LR problem: compare actual mole ratio of reactants to theoretical mole ratio.

29 29 2 Al + 3 Cl 2 ---> Al 2 Cl 6 Reactants must be in the mole ratio Step 1 of LR problem: compare actual mole ratio of reactants to theoretical mole ratio. mol Cl 2 mol Al = 3 2

30 30 Deciding on the Limiting Reactant If then there is not enough Al to use up all the Cl 2, and the limiting reagent is Al. Al. 2 Al + 3 Cl 2 Al 2 Cl 6 mol Cl Cl2 mol Al > 32

31 31 Deciding on the Limiting Reactant If then there is not enough Cl 2 to use up all the Al, and the limiting reagent is 2 Al + 3 Cl 2 ---> Al 2 Cl 6 Cl 2 mol Cl Cl2 mol Al < 32

32 32 We have 5.40 g of Al and 8.10 g of Cl 2 Step 2 of LR problem: Calculate moles of each reactant 5.40 g Al g Al 1 mol 27.0 g = mol Al mol Al 8.10 g Cl 2 1 mol 70.9 g = mol Cl mol Cl 2

33 33 Find mole ratio of reactants This should be 3/2 or 1.5/1 if reactants are present in the exact stoichiometric ratio. Limiting reagent is Cl 2 mol Cl Cl2 mol Al = mol mol =

34 34 Mix 5.40 g of Al with 8.10 g of Cl 2. What mass of Al 2 Cl 6 can form? Limiting reactant = Cl 2 Base all calculations on Cl 2 moles Cl 2 moles Al 2 Cl 6 grams Cl 2 grams Al 2 Cl 6

35 35 CALCULATIONS: calculate mass of Al 2 Cl 6 expected. Step 1: Calculate moles of Al 2 Cl 6 expected based on LR mol Cl Cl 2 1 mol Al Al2Cl6 3 mol Cl 2 = mol Al mol Al 2 Cl 6 Step 2: Calculate mass of Al 2 Cl 6 expected based on LR mol Al Al 2 Cl g Al 2Cl6 mol = 10.1 g Al 10.1 g Al 2 Cl 6

36 36 Cl 2 was the limiting reactant.Cl 2 was the limiting reactant. Therefore, Al was present in excess. Therefore, Al was present in excess. But how much? First find how much Al was required.First find how much Al was required. Then find how much Al is in excess.Then find how much Al is in excess. How much of which reactant will remain when reaction is complete?

37 37 2 Al + 3 Cl 2 products mol mol = LR Calculating Excess Al

38 38 2 Al + 3 Cl 2 products mol mol = LR Calculating Excess Al Excess Al = Al available - Al required = mol mol = mol Al in excess mol Cl 2 2 mol Al 3 mol Cl 2 = mol Al req' mol Al req'd

39 39 N I 2 2 NI 3 Nitrogen and iodine react to form nitrogen tri iodide. If 50.0 g of nitrogen is mixed with g iodine, calculate the number of grams of product formed and the grams of reactant remaining g/mol g/mol g/mol 1.79 mole 1.38 mole L.R. (0.460 S.U.) 50.0g -12.9g 0g left 363 g 37.1g left 350.0g g = 400.0g = 363g g

40 Using Stoichiometry to Determine a Formula Burn g of a hydrocarbon, C x H y, and produce g of CO 2 and g of H 2 O. What is the empirical formula of C x H y ? C x H y + oxygen g CO g H 2 O C x H y + oxygen g CO g H 2 O 40

41 41 Chemical Molecular Analysis in the lab in the lab

42 42 Using Stoichiometry to Determine a Formula First, recognize that all C in CO 2 and all H in H 2 O is from C x H y. 1.Calculate moles of C in CO x mol C 2.Calculate moles of H in H 2 O x mol H C x H y + some oxygen g CO g H 2 O

43 43 Using Stoichiometry to Determine a Formula Now find ratio of mol H/mol C to find values of x and y in C x H y x mol H/ 8.61 x mol C = 1.33 mol H / 1.00 mol C = 4 mol H / 3 mol C Empirical formula = C 3 H 4 C x H y + some oxygen g CO g H 2 O

44 44 Chemical Analysis Combustion - Determine a Formula Recognize that all C in CO 2 and all H in H 2 O is from C x H y. 1.Calculate moles of C in CO 2 C x H y + some oxygen g CO g H 2 O g CO 2 mole CO 2 mole C 44.0 g CO 2 mole CO g CO 2 mole CO 2 = mole C

45 45 Chemical Analysis Combustion - Determine a Formula 2. Calculate moles of H in H 2 O C x H y +some oxygen 0.379gCO g H 2 O g H 2 O mole H 2 O 2 mole H 18.0 g H 2 O mole H 2 O = mole H

46 46 Chemical Analysis Combustion - Determine a Formula Now find ratio of mol H/mol C to find values of x and y in C x H y mol H/ mol C = 1.34 mol H / 1.00 mol C = 4.02 mol H / 3.00 mol C Empirical formula = C 3 H 4 C x H y + some oxygen g CO g H 2 O

47 47 Sample Problems 1) A g sample of an unknown compound containing only carbon, hydrogen, and oxygen was burned to produced g of CO 2 and g of H 2 O. Determine the empirical formula. Given that the molecular weight is approximately 90 g/mole, determine the molecular formula g CO 2 mole CO 2 mole C 12.0 g C 44.0g CO 2 mole CO 2 mole C = g C g H 2 O mole H 2 O 2 mole H 1.0 g H 18.0 g H 2 O mole H 2 O mole H = g H

48 48 Sample Problems 1) A g sample of an unknown compound containing only carbon, hydrogen, and oxygen was burned to produced g of CO 2 and g of H 2 O. Determine the empirical formula. Given that the molecular weight is approximately 90 g/mole, determine the molecular formula g C g C g H g C, H, O g C, H, O g O g O

49 g 1 mole 12.0 g.070 g 1 mole 1.0 g C H O Empirical formula C 2 H 6 O..186 g 1 mole 16.0 g Sample Problems

50 50 Sample Problems Empirical formula C 2 H 6 O = 2 Molecular formula C 4 H 12 O 2 Alternate method. Use grams of cmpd. and its molar mass to determine moles of cmpd. Divide moles of C and H by these moles to find the subscripts for C and H. The subscript for O can be determinded by difference.

51 51 Sample Problems 2) A g sample of ascorbic acid, vitamin C, was burned to produce g of CO 2 and g H 2 O. Ascorbic acid contains only carbon, hydrogen, and oxygen. Determine the empirical formula. Given that the molecular weight is approximately 180 g/mole, determine the molecular formula g CO 2 mole CO 2 mole C 12.0 g C 44.0 g CO 2 mole CO 2 mole C g C = g C g H 2 O mole H 2 O 2 mole H 1.0 g H 18.0 g H 2 O mole H 2 O mole H g H = g H

52 52 Sample Problems 2) A g sample of ascorbic acid, vitamin C, was burned to produce g of CO 2 and g H 2 O. Ascorbic acid contains only carbon, hydrogen, and oxygen. Determine the empirical formula. Given that the molecular weight is approximately 180 g/mole, determine the molecular formula g C g H g C, H, O g O g O

53 g 1 mole 12.0 g.0057g 1 mole 1.0 g C H O Empirical formula C 3 H 4 O g 1 mole 16.0 g Sample Problems

54 54 Sample Problems Empirical formula C 3 H 4 O = 2 Molecular formula C 6 H 8 O 6

55 55 Chemical Analysis Mixtures - Determine a Percent 1. The amount of calcium present in milk can be determined by adding oxalate ion, C 2 O 4 2- (in the form of its water-soluble sodium salt, Na 2 C 2 O 4 ); the insoluble compound calcium oxalate is precipitated. Suppose you take a 75.0 g sample of milk and isolate g of calcium oxalate from it. What is the weight percentage of calcium in the milk? MILK (Ca 2+ ) CaC 2 O 4 Na 2 C 2 O 4

56 56 Chemical Analysis Mixtures - Determine a Percent MILK (Ca 2+ ) CaC 2 O g0.288 g ? % Ca Na 2 C 2 O 4 % Ca = g Ca 75.0 g milk X 100

57 57 Chemical Analysis Mixtures - Determine a Percent MILK (Ca 2+ ) CaC 2 O g0.288 g ? % Ca Na 2 C 2 O 4 % Ca = g Ca 75.0 g milk X g CaC 2 O 4 mole CaC 2 O 4 mole Ca 40.1 g Ca g CaC 2 O 4 mole CaC 2 O 4 mole Ca = g Ca = 0.120%

58 58 Chemical Analysis Mixtures - Determine a Percent 2. A 4.22 g sample of calcium chloride and sodium chloride was dissolved in water, and the solution was treated with sodium carbonate to precipitate the calcium as calcium carbonate. After isolating the solid calcium carbonate, it was heated to drive off the carbon dioxide and form g of calcium oxide. What is the weight percent of calcium chloride in the original 4.22 g sample?

59 59 Chemical Analysis Mixtures - Determine a Percent CaCl 2 /NaCl CaCO 3 Na 2 CO 3 CaO HeatHeat g 4.22 g % CaCl 2 = g CaCl g sample X 100

60 60 Chemical Analysis Mixtures - Determine a Percent g CaO mole CaO mole CaCl g CaCl g CaO mole CaO 4 mole CaCl 2 = 1.90 g CaCl 2 = 45.0% % CaCl 2 = 1.90 g CaCl g sample X 100

61 61 Practice Problems 1. Balance the following equations: CS 2 + Cl 2 --> CCl 4 + S 2 Cl 2 N 2 + O 2 --> NO C 8 H 18 + O 2 --> CO 2 + H 2 O 2. Write the formula equation for each of the following: sodium + water --> sodium hydroxide + hydrogen magnesium + oxygen --> magnesium oxide aluminum + hydrochloric acid --> aluminum chloride + hydrogen

62 62 Practice Problems 2. (continue) aluminum + hydrochloric acid --> aluminum chloride + hydrogen aluminum chloride + hydrogen sodium chlorate --> sodium chloride + oxygen mercury(II) sulfate + ammonium sulfide --> mercury(II) sulfide + ammonium sulfate mercury(II) sulfide + ammonium sulfate iron + cupric sulfate --> iron(III) sulfate + copper

63 63 Practice Problems For problems H 2 + N 2  2 NH 3 3. How many moles of H 2 are required to react 4.2 moles of N 2 ? 4. How many moles of H 2 are required to react 74 grams of N 2 ? 5. How many grams of NH 3 would be produced from 45 g of H 2 ? 6. How many moles of NH 3 would be produced from the reaction of 18.5 g H 2 and 95 g of N 2 ?

64 64 Practice Problems 7. Phosphine, PH 3, is formed when calcium phosphide is added to water. How many grams of phosphine can be obtained from 205 g of calcium phosphide? 8. How many grams of iron will be required to release all of the antimony from 10.0 g antimony trisulfide? (Ferrous sulfide is formed) 9. If calcium oxide were prepared by heating calcium carbonate, how many grams of the carbonate would be required to produce the 15.0 g of the oxide?

65 65 Practice Problems 10. How many grams of cupric sulfate are needed to completely react with 145 g of sodium chloride? How many grams of sodium sulfate could be produced by this reaction? 11. How many grams of sulfuric acid will react with 40.0 g of aluminum metal? grams of zinc are added to 120 grams of nitrous acid. How many grams of hydrogen gas are evolved?

66 66 Practice Problems grams of hydrogen and 75 grams of oxygen are exploded together in a reaction tube. How many grams of water are produced? What other gas is found in the tube(besides water vapor) after the reaction, and how many grams of this gas are there?

67 67 Practice Problems 14. A white powder was a mixture of NaBr and NaNO 3. A sample of the powder weighing g was dissolved in water and a solution of AgNO 3 was added until the precipitation of AgBr was complete. The reaction mixture was filtered and dried and the precipitate of AgBr weighed g. What was the percentage by weight of NaBr in the original sample?

68 68 Practice Problems 15. A 4.81 g sample of an unknown compound containing only carbon, hydrogen, and nitrogen was burned to produce g of CO 2 and 1.68 g of H 2 O. Determine the empirical formula.

69 69 Practice Problems Answers 1. 1,3,1,11,1,22,25,16, Na + 2 HOH --> 2 NaOH + H 2 2 Mg + O 2 --> 2 MgO 2 Al + 6 HCl --> 2 AlCl H 2 2 NaClO 3 --> 2 NaCl + 3 O 2 HgSO 4 + (NH 4 ) 2 S --> Hg + (NH 4 ) 2 SO4 2 Fe + 3 CuSO 4 --> Fe 2 (SO 4 ) Cu mole H mole H g NH mole NH 3

70 70 Practice Problems Answers g PH g Fe g CaCO g CuSO4, 176 g Na 2 SO g H 2 SO g H g H 2 O, 0.6 g H % 15. C 5 H 3 N

71 71 Writing Equations zinc + chlorine ---> zinc chloride Zn (s) + Cl 2 (g) --> ZnCl 2 (s) Combination, Synthesis

72 72 Writing Equations potassium nitrate --> potassium nitrite + oxygen potassium nitrate --> potassium nitrite + oxygen KNO 3 (s) --> KNO 2 (s) + O 2 (g) 2 KNO 3 (s) --> 2 KNO 2 (s) + O 2 (g) 2 KNO 3 (s) --> 2 KNO 2 (s) + O 2 (g) Decomposition

73 73 Writing Equations magnesium bromide + chlorine --> magnesium chloride + bromine MgBr 2 (s) + Cl 2 (g) --> MgCl 2 (s) + Br 2 (g) Single Displacement

74 74 Writing Equations calcium hydroxide + hydrochloric acid --> calcium chloride + water Ca(OH) 2 (aq) + HCl (aq) --> CaCl 2 (aq) + H 2 O (l) Ca(OH) 2 (aq) + 2HCl (aq) --> CaCl 2 (aq) + 2 H 2 O (l) Double Displacement

75 75 Writing Equations zinc chloride + ammonium sulfide --> zinc sulfide + ammonium chloride ZnCl 2 (aq) + (NH 4 ) 2 S (aq) --> ZnS (s) + NH 4 Cl (aq) ZnCl 2 (aq) + (NH 4 ) 2 S (aq) --> ZnS (s) + 2 NH 4 Cl (aq)

76 76 Writing Equations aluminum + cupric chloride --> copper + aluminum chloride Al (s) + CuCl 2 (aq) --> Cu (s) + AlCl 3 (aq) Al (s) + CuCl 2 (aq) --> Cu (s) + AlCl 3 (aq) 2 Al (s) + 3 CuCl 2 (aq) --> 3 Cu (s) + 2 AlCl 3 (aq) 2 Al (s) + 3 CuCl 2 (aq) --> 3 Cu (s) + 2 AlCl 3 (aq)

77 77 PROBLEM: How many moles of H 2 are required to produce 9 moles of NH 3 ? STEP 1 Write the balanced chemical equation 3 H 2 + N 2 --> 2 NH 3 3 H 2 + N 2 --> 2 NH 3

78 78 STEP 2 Write the given and requested information below the equation. 3 H 2 + N 2 --> 2 NH 3 3 H 2 + N 2 --> 2 NH 3 ? mole 9 mole ? mole 9 mole

79 79 3 H 2 + N 2 --> 2 NH 3 ? mole 9 mole STEP 3 Calculate using the information. = 10 mole H 2 9 mole NH 3 3 mole H 2 2 mole NH 3 2 mole NH 3

80 80 PROBLEM: How many moles of NH 3 can be produced from 10.4 moles of N 2 ? 3 H 2 + N 2 --> 2 NH mole? mole = 20.8 mole NH mole N 2 2 mole NH 3 mole N 2 mole N 2

81 81 PROBLEM: How many grams of H 2 are required to produce 8 moles of NH 3 ? 3 H 2 + N 2 --> 2 NH 3 ? g8 mole 8 mole NH 3 3 mole H g H 2 2 mole NH 3 mole H 2 2 mole NH 3 mole H 2 = 20 g H 2

82 82 PROBLEM: How many moles of NH 3 can be produced from 55 grams of N 2 ? 3 H 2 + N 2 --> 2 NH 3 55 g? mole 55 g N 2 1 mole N 2 2 mole NH 3 55 g N 2 1 mole N 2 2 mole NH g N 2 mole N g N 2 mole N 2 = 3.9 mole NH 3

83 83 PROBLEM: How many grams of H 2 are required to react 24 grams of N 2 ? 3 H 2 + N 2 --> 2 NH 3 ? g 24 g 24 g N 2 mole N 2 3 mole H g H g N 2 mole N 2 mole H g N 2 mole N 2 mole H 2 = 5.1 g H 2

84 84 PROBLEM: How many grams of N 2 are required to produce 155 grams of NH 3 ? 3 H 2 + N 2 --> 2 NH 3 ? g 155 g 155 g NH 3 mole NH 3 mole N g N g NH 3 2 mole NH 3 mole N g NH 3 2 mole NH 3 mole N 2 = 128 g N 2

85 85 Sample Problems 1) Sulfur dioxide may be oxidized to sulfur trioxide. How many grams of sulfur dioxide could be converted by this process if g of oxygen are available for the oxidation? 2 SO 2 + O 2 --> 2 SO 3 ? g g ? g g g O 2 mole O 2 2 mole SO g SO g O 2 mole O 2 mole SO g O 2 mole O 2 mole SO 2 = 401 g SO 2

86 86 Sample Problems 2) Lightning discharges in the atmosphere catalyze the conversion of nitrogen to nitrogen dioxide. How many grams of nitrogen would be required to make 25.0 g of nitrogen dioxide in this way? N O 2 --> 2 NO 2 N O 2 --> 2 NO 2 ? g 25.0 g ? g 25.0 g 25.0 g NO 2 mole NO 2 1 mole N g N g NO 2 2 mole NO 2 mole N g NO 2 2 mole NO 2 mole N 2 = 7.61 g N 2

87 87 Sample Problems 3) Ferric oxide may be reduced to pure iron with coke (pure carbon). Suppose that g of ferric oxide is available. How many grams of carbon would be needed? 2 Fe 2 O C 4 Fe + 3 CO g 2 Fe 2 O C 4 Fe + 3 CO g ? g g Fe 2 O 3 mole Fe 2 O 3 3 mole C 12.0 g C g Fe 2 O 3 2 mole Fe 2 O 3 mole C g Fe 2 O 3 2 mole Fe 2 O 3 mole C = 16.9 g C

88 88 Sample Problems 4) Zinc metal will react with hydrochloric acid to produce hydrogen gas. If 50.0 g of zinc is to be used in the reaction, how many grams of acid would be needed to completely react with all of the zinc? How many grams of hydrogen gas would be produced? Zn + 2 HCl --> ZnCl 2 + H 2 Zn + 2 HCl --> ZnCl 2 + H g ? g ? g 50.0 g ? g ? g 50.0 g Zn mole Zn 2 mole HCl 36.5 g HCl 65.4 g Zn mole Zn mole HCl 65.4 g Zn mole Zn mole HCl = 55.8gHCl

89 89 Sample Problems 4) Zinc metal will react with hydrochloric acid to produce hydrogen gas. If 50.0 g of zinc is to be used in the reaction, how many grams of acid would be needed to completely react with all of the zinc? How many grams of hydrogen gas would be produced? Zn + 2 HCl --> ZnCl 2 + H 2 Zn + 2 HCl --> ZnCl 2 + H g ? g ? g 50.0 g ? g ? g 50.0 g Zn mole Zn mole H g H g Zn mole Zn mole H g Zn mole Zn mole H 2 = 1.5 g H 2

90 90 Sample Problems 5) Phosphoric acid, H 3 PO 4, is produced in the reaction between calcium phosphate and sulfuric acid. How many grams of phosphoric acid would be produced from 55 grams of calcium phosphate? What other product is formed and in what quantity? Ca 3 (PO 4 ) H 2 SO 4 --> 3 CaSO H 3 PO 4 55 g ? g ? g ? g ? g 55 g Ca 3 (PO4) 2 mole Ca 3 (PO4) 2 3 mole CaSO g CaSO gCa 3 (PO4) 2 mole Ca 3 (PO4) 2 mole CaSO gCa 3 (PO4) 2 mole Ca 3 (PO4) 2 mole CaSO 4 = 72 g CaSO 4

91 91 Sample Problems produced 5) Phosphoric acid, H 3 PO 4, is produced in the reaction between calcium phosphate and sulfuric acid. How many grams of phosphoric acid would be produced from 55 grams of calcium phosphate? What other product is formed and in what quantity? Ca 3 (PO 4 ) H 2 SO 4 --> 3 CaSO H 3 PO 4 55 g ? g ? g ? g ? g 55 g Ca 3 (PO4) 2 mole Ca 3 (PO4) 2 2 mole H 3 PO g H 3 PO gCa 3 (PO4) 2 mole Ca 3 (PO4) 2 mole H 3 PO gCa 3 (PO4) 2 mole Ca 3 (PO4) 2 mole H 3 PO 4 = 35 g H 3 PO 4

92 92 Percent Yield % yield= actual yield theoretical yield 100%

93 93 PROBLEM: If 19.3 g H 2 produces 78.5 g NH 3, what is the percent yield? 3 H 2 + N 2 --> 2 NH g ? g 19.3 g H 2 mole H 2 2 mole NH g NH g H 2 3 mole H 2 mole NH g H 2 3 mole H 2 mole NH 3 = 110 g NH 3 % yield = 78.5g 110 g x 100 = 71 %

94 94 PROBLEM: If the yield obtained is 75%, how many grams of NH 3 would be obtained from 10.4 g of N 2 ? 3 H 2 + N 2 --> 2 NH g ? g 10.4 g N 2 mole N 2 2 mole NH g NH g N 2 mole N 2 mole NH g N 2 mole N 2 mole NH 3 = 12.6 g NH g NH 3 x.75 = 9.4 g NH 3

95 95 PROBLEM: 75.0 grams of potassium hydroxide are permitted to react with 50.0 grams of hydrochloric acid. How many grams of potassium chloride are formed? STEP 1 Write the balanced chemical equation KOH + HCl --> KCl + HOH KOH + HCl --> KCl + HOH

96 96 STEP 2 Write the given and requested information below the equation. KOH + HCl --> KCl + HOH KOH + HCl --> KCl + HOH 75.0 g 50.0 g ?g 75.0 g 50.0 g ?g

97 97 KOH + HCl --> KCl + HOH KOH + HCl --> KCl + HOH 75.0 g 50.0 g ?g 75.0 g 50.0 g ?g STEP 3 Calculate the product assuming that each reactant is the limiting reagent g KOH mole KOH mole KCl 74.6 g KCl 56.1 g KOH mole KOH mole KCl 56.1 g KOH mole KOH mole KCl = 99.7g KCl 50.0 g HCl mole HCl mole KCl 74.6 g KCl 36.5 g HCl mole HCl mole KCl 36.5 g HCl mole HCl mole KCl = 102 g KCl

98 98 KOH + HCl --> KCl + HOH KOH + HCl --> KCl + HOH 75.0 g 50.0 g ?g 75.0 g 50.0 g ?g STEP 4 Determine the limiting reactant and the actual amount of product g KOH can produce 99.7 g KCl KOH is the limiting reactant, HCl is the excess reactant g HCl can produce 102 g KCl 99.7 g KCl

99 99 KOH + HCl --> KCl + HOH KOH + HCl --> KCl + HOH 75.0 g 50.0 g ?g 75.0 g 50.0 g ?g STEP 5 Determine the amount of excess reactant by calculating the amount used and subtracting from the starting amount g KOH mole KOH mole HCl 36.5 g HCl 56.1 g KOH mole KOH mole HCl 56.1 g KOH mole KOH mole HCl = 48.8g HCl used used 50.0 g = 1.2 g HCl left

100 100 Sample Problems 1) How many grams of carbon dioxide can be obtained from the action of grams of hydrobromic acid on grams of calcium carbonate? 2 HBr + CaCO 3 --> HOH + CO 2 + CaBr g g ? g g g ? g g HBr mole HBr mole CO g CO g HBr 2 mole HBr mole CO g HBr 2 mole HBr mole CO 2 = 27.2 g CO g CaCO 3 mole CaCO 3 mole CO g CO g CaCO 3 mole CaCO 3 mole CO g CaCO 3 mole CaCO 3 mole CO 2 = 44.0 g CO 2

101 101 Sample Problems 2) How many grams of ammonia are evolved when 34 grams of ammonium chloride are added to 140 grams of barium hydroxide? 2 NH 4 Cl + Ba(OH) 2 --> BaCl NH 3 +2 HOH 34 g 140 g ? g 34 g 140 g ? g 34 g NH 4 Cl mole NH 4 Cl 2 mole NH g NH g NH 4 Cl 2 mole NH 4 Cl mole NH g NH 4 Cl 2 mole NH 4 Cl mole NH 3 = 11 g NH g Ba(OH) 2 mole Ba(OH) 2 2 mole NH g NH g Ba(OH) 2 mole Ba(OH) 2 mole NH g Ba(OH) 2 mole Ba(OH) 2 mole NH 3 = 28.0 gNH 3

102 102 Sample Problems 3) grams of lithium metal is dropped into liter of water. How many grams of hydrogen are produced? 2 Li + 2 HOH --> 2 LiOH + H g g ? g g g ? g g Li mole Li mole H g H g Li 2 mole Li mole H 2 = 14 g H g HOH mole HOH mole H g H g HOH 2 mole HOH mole H 2 = 56 g H 2

103 103 Sample Problems 4).50.0 grams of oxygen are available for the combustion of 25.0 grams of carbon. How many grams in excess is the oxygen or carbon ? O 2 + C --> CO 2 O 2 + C --> CO 2 ? g excess 50.0 g 25.0 g ? g excess 50.0 g 25.0 g 50.0 g O 2 mole O 2 mole C 12.0 g C 32.0 g O 2 mole O 2 mole C 32.0 g O 2 mole O 2 mole C = 18.8g C used 25.0 g = 6.2 g C left

104 104 Sample Problems 5) grams of sulfuric acid is added to grams of barium peroxide. Which reactant is in excess and by how many grams? H 2 SO 4 + BaO 2 --> BaSO 4 + H 2 O 2 H 2 SO 4 + BaO 2 --> BaSO 4 + H 2 O 2 g excess 140. g 230. g 140.gH 2 SO 4 mole H 2 SO 4 mole BaO g BaO g H 2 SO 4 mole H 2 SO 4 mole BaO 2 = 242 g BaO 2 Since only 230. g available, BaO 2 is the limiting reactant

105 105 Sample Problems 5) grams of sulfuric acid is added to grams of barium peroxide. Which reactant is in excess and by how many grams? H 2 SO 4 + BaO 2 --> BaSO 4 + H 2 O 2 H 2 SO 4 + BaO 2 --> BaSO 4 + H 2 O 2 g excess 140. g 230. g 230. g BaO 2 mole BaO 2 mole H 2 SO g H 2 SO g BaO 2 mole BaO 2 mole H 2 SO 4 = 133 g H 2 SO 4 used 140. g = 7 g H 2 SO 4 left


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