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Chapter 12 Stoichiometry u Greek for measuring elements u The calculations of quantities in chemical reactions based on a balanced equation. u We can.

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Presentation on theme: "Chapter 12 Stoichiometry u Greek for measuring elements u The calculations of quantities in chemical reactions based on a balanced equation. u We can."— Presentation transcript:

1

2 Chapter 12

3 Stoichiometry u Greek for measuring elements u The calculations of quantities in chemical reactions based on a balanced equation. u We can interpret balanced chemical equations several ways.

4 In terms of Particles u Atom - Element u Molecule Molecular compound (non- metals) or diatomic (O 2 etc.) u Formula unit Ionic Compounds (Metal and non- metal)

5 2H 2 + O 2 2H 2 O u Two molecules of hydrogen and one molecule of oxygen form two molecules of water. 2 Al 2 O 3 Al + 3O 2 2formula unitsAl 2 O 3 form4 atoms Al and3moleculesO2O2

6 2Na + 2H 2 O 2NaOH + H 2 2atomsNaform 2 formula units NaOHand1moleculeH2H2 and2moleculesH2OH2O

7 Look at it differently 2H 2 + O 2 2H 2 O u 2 dozen molecules of hydrogen and 1 dozen molecules of oxygen form 2 dozen molecules of water. u 2 x (6.02 x ) molecules of hydrogen and 1 x (6.02 x ) molecules of oxygen form 2 x (6.02 x ) molecules of water. u 2 moles of hydrogen and 1 mole of oxygen form 2 moles of water.

8 In terms of Moles 2 Al 2 O 3 Al + 3O 2 u The coefficients tell us how many moles of each kind

9 36.04 g reactants In terms of mass u The law of conservation of mass applies u We can check using moles 2H 2 + O 2 2H 2 O 2 moles H g H 2 1 moles H 2 =4.04 g H 2 1 moles O g O 2 1 moles O 2 =32.00 g O 2

10 In terms of mass 2H 2 + O 2 2H 2 O 2 moles H 2 O g H 2 O 1 mole H 2 O = g H 2 O 2H 2 + O 2 2H 2 O g (H 2 + O 2 ) =36.04 g H 2 O

11 Mole to mole conversions 2 Al 2 O 3 Al + 3O 2 u every time we use 2 moles of Al 2 O 3 we make 3 moles of O 2 2 moles Al 2 O 3 3 mole O 2 or 2 moles Al 2 O 3 3 mole O 2

12 Mole to Mole conversions u How many moles of O 2 are produced when 3.34 moles of Al 2 O 3 decompose? 2 Al 2 O 3 Al + 3O moles Al 2 O 3 2 moles Al 2 O 3 3 mole O 2 =5.01 moles O 2

13 Practice 2C 2 H O 2 4CO H 2 O u If 3.84 moles of C 2 H 2 are burned, how many moles of O 2 are needed?

14 Practice 2C 2 H O 2 4CO H 2 O u How many moles of C 2 H 2 are needed to produce 8.95 mole of H 2 O?

15 Mass in Chemical Reactions How much do you make? How much do you need?

16 We cant measure moles!! u What can we do? u We can convert grams to moles. Periodic Table u Then use moles to change chemicals Balanced equation u Then turn the moles back to grams. Periodic table

17 Periodic Table Moles A Moles B Mass g B Periodic Table Balanced Equation Mass g A Decide where to start based on the units you are given and stop based on what unit you are asked for

18 Conversions 2C 2 H O 2 4CO H 2 O u How many moles of C 2 H 2 are needed to produce 8.95 g of H 2 O?

19 Conversions 2C 2 H O 2 4CO H 2 O u If 2.47 moles of C 2 H 2 are burned, how many g of CO 2 are formed?

20 For example... u If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how much solid copper would form? Fe + CuSO 4 Fe 2 (SO 4 ) 3 + Cu 2Fe + 3CuSO 4 Fe 2 (SO 4 ) 3 + 3Cu 10.1 g Fe g Fe 1 mol Fe 2 mol Fe 3 mol Cu 1 mol Cu g Cu =17.2 g Cu

21 2Fe + 3CuSO 4 Fe 2 (SO 4 ) 3 + 3Cu mol Fe 2 mol Fe 3 mol Cu = mol Cu 1 mol Cu g Cu =17.3 g Cu

22 Could have done it 10.1 g Fe g Fe 1 mol Fe 2 mol Fe 3 mol Cu 1 mol Cu g Cu =17.3 g Cu

23 u To make silicon for computer chips they use this reaction SiCl 4 + 2Mg 2MgCl 2 + Si u How many moles of Mg are needed to make 9.3 g of Si?

24 u To make silicon for computer chips they use this reaction SiCl 4 + 2Mg 2MgCl 2 + Si u 3.74 g of Mg would make how many moles of Si?

25 u To make silicon for computer chips they use this reaction SiCl 4 + 2Mg 2MgCl 2 + Si u How many grams of MgCl 2 are produced along with 9.3 g of silicon?

26 u The U. S. Space Shuttle boosters use this reaction 3 Al(s) + 3 NH 4 ClO 4 Al 2 O 3 + AlCl NO + 6H 2 O u How much Al must be used to react with 652 g of NH 4 ClO 4 ? u How much water is produced? u How much AlCl 3 ?

27 How do you get good at this?

28 Gases and Reactions

29 We can also change u Liters of a gas to moles u At STP 0ºC and 1 atmosphere pressure u At STP 22.4 L of a gas = 1 mole

30 For Example u If 6.45 grams of water are decomposed, how many liters of oxygen will be produced at STP? H 2 O H 2 + O 2 2H 2 O 2H 2 + O g H 2 O g H 2 O 1 mol H 2 O 2 mol H 2 O 1 mol O L O 2

31 Your Turn u How many liters of CO 2 at STP will be produced from the complete combustion of 23.2 g C 4 H 10 ?

32 Example u How many liters of CH 4 at STP are required to completely react with 17.5 L of O 2 ? CH 4 + 2O 2 CO 2 + 2H 2 O 17.5 L O2O L O2O2 1 mol O2O2 2 O2O2 1 CH 4 1 mol CH L CH 4 = 8.75 L CH L O 2 1 mol O 2 1 mol CH L CH 4

33 Avogadro told us u Equal volumes of gas, at the same temperature and pressure contain the same number of particles. u Moles are numbers of particles u You can treat reactions as if they happen liters at a time, as long as you keep the temperature and pressure the same.

34 Example u How many liters of CO 2 at STP are produced by completely burning 17.5 L of CH 4 ? CH 4 + 2O 2 CO 2 + 2H 2 O 17.5 L CH 4 1 L CH 4 1 L CO 2 = 17.5 L CO 2

35 Particles u We can also change between particles and moles. u 6.02 x Molecules Atoms Formula units

36 Example u If 2.8 g of C 4 H 10 are burned completely, how many water molecules will be made?

37 particles A Liters A grams A particles B Liters B grams Bmoles Amoles B 22.4 L 6.02 x PT equation

38 Limiting Reagent u If you are given one dozen loaves of bread, a gallon of mustard and three pieces of salami, how many salami sandwiches can you make? u The limiting reagent is the reactant you run out of first. u The excess reagent is the one you have left over. u The limiting reagent determines how much product you can make

39 How do you find out? u Do two stoichiometry problems. u The one that makes the least product is the limiting reagent. u For example u Copper reacts with sulfur to form copper ( I ) sulfide. If 10.6 g of copper reacts with 3.83 g S how much product will be formed?

40 u If 10.6 g of copper reacts with 3.83 g S. How many grams of product will be formed? 2Cu + S Cu 2 S 10.6 g Cu 63.55g Cu 1 mol Cu 2 mol Cu 1 mol Cu 2 S 1 mol Cu 2 S g Cu 2 S = 13.3 g Cu 2 S 3.83 g S 32.06g S 1 mol S 1 S 1 Cu 2 S 1 mol Cu 2 S g Cu 2 S = 19.0 g Cu 2 S = 13.3 g Cu 2 S Cu is Limiting Reagent

41 How much excess reagent? u Use the limiting reagent to find out how much excess reagent you used u Subtract that from the amount of excess you started with

42 Mg(s) +2 HCl(g) MgCl 2 (s) +H 2 (g) u If 4.87 mol of magnesium and 9.84 mol of HCl gas are reacted, how many moles of gas will be produced? u What is the limiting reagent?

43 Mg(s) +2 HCl(g) MgCl 2 (s) +H 2 (g) u If 4.87 mol of magnesium and 9.84 mol of HCl gas are reacted, how many moles of gas will be produced? u How much excess reagent remains?

44 u If 10.3 g of aluminum are reacted with 51.7 g of CuSO 4 how much copper will be produced? u How much excess reagent will remain?

45

46 Yield u The amount of product made in a chemical reaction. u There are three types u Actual yield- what you get in the lab when the chemicals are mixed u Theoretical yield- what the balanced equation tells you you should make. u Percent yield u Percent yield = Actual x 100 % Theoretical

47 Example u 6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate. 2Al + 3 CuSO 4 Al 2 (SO 4 ) 3 + 3Cu u What is the actual yield? u What is the theoretical yield? u What is the percent yield? u If you had started with 9.73 g of Al, how much copper would you expect?

48 Details u Percent yield tells us how efficient a reaction is. u Percent yield can not be bigger than 100 %.

49 Energy in Chemical Reactions How Much? In or Out?

50 49 Energy u Energy is measured in Joules or calories u Every reaction has an energy change associated with it u Exothermic reactions release energy, usually in the form of heat. u Endothermic reactions absorb energy u Energy is stored in bonds between atoms u Making bonds gives energy u Breaking bonds takes energy

51 50 In terms of bonds C O O C O O Breaking this bond will require energy C O O O O C Making these bonds gives you energy In this case making the bonds gives you more energy than breaking them

52 51 Exothermic u The products are lower in energy than the reactants u Releases energy u Often release heat

53 52 C + O 2 CO 2 Energy ReactantsProducts C + O 2 CO kJ kJ

54 53 Endothermic u The products are higher in energy than the reactants u Absorbs energy u Absorb heat

55 54 CaCO 3 CaO + CO 2 Energy ReactantsProducts CaCO 3 CaO + CO kJ CaCO kJ CaO + CO 2

56 55 Chemistry Happens in u MOLES u An equation that includes energy is called a thermochemical equation CH O 2 CO H 2 O kJ u 1 mole of CH 4 makes kJ of energy. u When you make kJ you make 2 moles of water

57 56 CH O 2 CO H 2 O kJ u If grams of CH 4 are burned completely, how much heat will be produced? g CH g CH 4 1 mol CH kJ =515 kJ

58 57 CH O 2 CO H 2 O kJ u How many liters of O 2 at STP would be required to produce 23 kJ of heat? u How many grams of water would be produced with 506 kJ of heat?

59 Calorimetry u Measuring heat. u Use a calorimeter. u Two kinds u Constant pressure calorimeter (called a coffee cup calorimeter) u An insulated cup, full of water. u The specific heat of water is 1 cal/gºC heat = specific heat x m x T

60 Example u A chemical reaction is carried out in a coffee cup calorimeter. There are 75.8 g of water in the cup, and the temperature rises from 16.8 ºC to 34.3 ºC. How much heat was released?

61 Calorimetry u Second type is called a bomb calorimeter. u Material is put in a container with pure oxygen. Wires are used to start the combustion. The container is put into a container of water. u The heat capacity of the calorimeter is known and/or tested. (cal/ ºC) u Multiply temperature change by the heat capacity to find heat

62 Bomb Calorimeter u thermometer u stirrer u full of water u ignition wire u Steel bomb u sample

63

64 Heats of Reaction

65 64 Enthalpy u The heat content a substance has at a given temperature and pressure u Cant be measured directly because there is no set starting point u The reactants start with a heat content u The products end up with a heat content u So we can measure how much heat content changes u Called change in enthalpy

66 65 Enthalpy u Symbol is H Change in enthalpy is H u delta H u If heat is released the heat content of the products is lower H is negative (exothermic) u If heat is absorbed the heat content of the products is higher H is positive (endothermic)

67 66 Energy ReactantsProducts Change is down H is <0

68 67 Energy ReactantsProducts Change is up H is > 0

69 68 Heat of Reaction u The heat that is released or absorbed in a chemical reaction Equivalent to H C + O 2 (g) CO 2 (g) kJ C + O 2 (g) CO 2 (g) H = kJ u In thermochemical equation it is important to say what state H 2 (g) + ½ O 2 (g) H 2 O(g) H = kJ H 2 (g) + ½ O 2 (g) H 2 O(l) H = kJ

70 69 Heat of Combustion u The heat from the reaction that completely burns 1 mole of a substance C 2 H O 2 2 CO H 2 O C 2 H 6 + O 2 CO 2 + H 2 O 2 C 2 H O 2 2 CO H 2 O C 2 H 6 + (7/2) O 2 CO H 2 O

71 70 Standard Heat of Formation The H for a reaction that produces 1 mol of a compound from its elements at standard conditions u Standard conditions 25°C and 1 atm. u Symbol is u The standard heat of formation of an element is 0 u This includes the diatomics

72 71 What good are they? u There are tables (pg. 190) of heats of formations u For most compounds it is negative Because you are making bonds Making bonds is exothermic u The heat of a reaction can be calculated by subtracting the heats of formation of the reactants from the products

73 72 Examples CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(g) CH 4 (g) = kJO 2 (g) = 0 kJCO 2 (g) = kJH 2 O(g) = kJ H= [ (-241.8)]-[ (0)] H= kJ

74 73 Examples 2 SO 3 (g) 2SO 2 (g) + O 2 (g)

75 74 Why Does It Work? If H 2 (g) + 1/2 O 2 (g) H 2 O(l) H= kJ then H 2 O(g) H 2 (g) + 1/2 O 2 (l) H = kJ u If you turn an equation around, you change the sign 2 H 2 O(g) H 2 (g) + O 2 (l) H = kJ u If you multiply the equation by a number, you multiply the heat by that number.

76 75 Why does it work? u You make the products, so you need their heats of formation u You unmake the reactants so you have to subtract their heats.

77 Energy ReactantsProducts reactants products elements

78 Energy ReactantsProducts reactants products elements


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