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Love, Life and Stoichiometry Stoichiometry Greek for “measuring elements” Greek for “measuring elements” The calculations of quantities in chemical reactions.

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Presentation on theme: "Love, Life and Stoichiometry Stoichiometry Greek for “measuring elements” Greek for “measuring elements” The calculations of quantities in chemical reactions."— Presentation transcript:

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2 Love, Life and Stoichiometry

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

4 In terms of Particles Element- atoms Element- atoms Molecular compound (non- metals)- molecule Molecular compound (non- metals)- molecule Ionic Compounds (Metal and non-metal) - formula unit Ionic Compounds (Metal and non-metal) - formula unit

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

6 Look at it differently 2H 2 + O 2   2H 2 O 2H 2 + O 2   2H 2 O 2 dozen molecules of hydrogen and 1 dozen molecules of oxygen form 2 dozen molecules of water. 2 dozen molecules of hydrogen and 1 dozen molecules of oxygen form 2 dozen molecules of water. 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. 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. 2 moles of hydrogen and 1 mole of oxygen form 2 moles of water. 2 moles of hydrogen and 1 mole of oxygen form 2 moles of water.

7 In terms of Moles 2 Al 2 O 3  Al + 3O 2 2 Al 2 O 3  Al + 3O 2 2Na + 2H 2 O  2NaOH + H 2 2Na + 2H 2 O  2NaOH + H 2 The coefficients tell us how many moles of each kind The coefficients tell us how many moles of each kind

8 In terms of mass The law of conservation of mass applies The law of conservation of mass applies We can check using moles We can check using moles 2H 2 + O 2   2H 2 O 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 = g O g H g Reactants

9 Therefore… 2H 2 + O 2   2H 2 O 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 No mass has been created or destroyed

10 Your turn Show that the following equation follows the Law of conservation of mass. Show that the following equation follows the Law of conservation of mass. 2 Al 2 O 3  Al + 3O 2 2 Al 2 O 3  Al + 3O 2

11 Mole to mole conversions 2 Al 2 O 3  Al + 3O 2 2 Al 2 O 3  Al + 3O 2 every time we use 2 moles of Al 2 O 3 we make 3 moles of O 2 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 How many moles of O 2 are produced when 3.34 moles of Al 2 O 3 decompose? 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 2 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 Your Turn 2C 2 H O 2  4CO H 2 O 2C 2 H O 2  4CO H 2 O If 3.84 moles of C 2 H 2 are burned, how many moles of O 2 are needed? If 3.84 moles of C 2 H 2 are burned, how many moles of O 2 are needed? How many moles of C 2 H 2 are needed to produce 8.95 mole of H 2 O? How many moles of C 2 H 2 are needed to produce 8.95 mole of H 2 O? If 2.47 moles of C 2 H 2 are burned, how many moles of CO 2 are formed? If 2.47 moles of C 2 H 2 are burned, how many moles of CO 2 are formed?

14 How do you get good at this?

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16 Stoichiometry Practice

17 Given the following equation: 2 KClO 3  2KCl + 3 O 2 How many moles of O 2 can be produced by letting moles of KClO 3 react?

18 2 KClO 3  2KCl + 3 O 2 The strategy will be to convert to. Remember, conversion factors are what you over what you. moles of O 2 want moles of KClO 3 have

19 KClO 3  2KCl + O moles of KClO 3 moles of O 2 moles of KClO 3 23 = 12 x (3/2) =

20 If ammonia, NH 3, is burned in air, the following reaction takes place: 4NH 3 + 3O 2  2 N 2 + 6H 2 O Given that you started with 51.0 g of NH 3, how many grams of water will be produced?

21 4NH 3 + 3O 2  2 N 2 + 6H 2 O The overall strategy will be to convert the of ammonia of ammonia, then convert the of ammonia of water, and then finally convert the of water of water. grams to moles moles to moles molesto grams

22 4NH 3 + 3O 2  2 N 2 + 6H 2 O grams NH 3 moles NH 3 moles H 2 O grams of H 2 O 51.0 g NH 3 1 mole NH 3 17 g NH 3 6 mole H 2 O 4 mole NH 3 18 g H 2 O 1 mole H 2 O = 81 g of H 2 O 

23 20.0 g of silver(I)nitrate is reacted with an excess of sodium choride to produce silver(I) chloride AgNO 3 + NaCl => AgCl + NaNO 3 What mass of silver(I) chloride is produced?

24 AgNO 3 + NaCl => AgCl + NaNO 3 The overall strategy will be to convert the of silver nitrate of silver nitrate, then convert the of silver nitrate of silver chloride, and then finally convert the of silver chloride of silver chloride. grams to moles moles to moles moles to grams

25 AgNO 3 + NaCl => AgCl + NaNO g AgNO 3 1 mole AgNO g AgNO 3 1 mole AgCl 1 mole AgNO g AgCl 1 mole AgCl = 16.9  grams AgNO 3 moles AgNO 3 moles AgCl grams AgCl g AgCl

26 Now go to mass to mass problems

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

28 We can’t measure moles!! What can we do? What can we do? We can convert grams to moles. We can convert grams to moles. Periodic Table Periodic Table Then do the math with the moles. Then do the math with the moles. Balanced equation Balanced equation Then turn the moles back to grams. Then turn the moles back to grams. Periodic table Periodic table

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

30 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

31 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

32 More Examples To make silicon for computer chips they use this reaction To make silicon for computer chips they use this reaction SiCl 4 + 2Mg  2MgCl 2 + Si SiCl 4 + 2Mg  2MgCl 2 + Si How many grams of Mg are needed to make 9.3 g of Si? How many grams of Mg are needed to make 9.3 g of Si? How many grams of SiCl 4 are needed to make 9.3 g of Si? How many grams of SiCl 4 are needed to make 9.3 g of Si? How many grams of MgCl 2 are produced along with 9.3 g of silicon? How many grams of MgCl 2 are produced along with 9.3 g of silicon?

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

34 We can also change Liters of a gas to moles Liters of a gas to moles At STP At STP –25ºC and 1 atmosphere pressure At STP 22.4 L of a gas = 1 mole At STP 22.4 L of a gas = 1 mole If 6.45 moles of water are decomposed, how many liters of oxygen will be produced at STP? If 6.45 moles of water are decomposed, how many liters of oxygen will be produced at STP?

35 For Example If 6.45 grams of water are decomposed, how many liters of oxygen will be produced at STP? 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 H 2 O  H 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

36 Your Turn How many liters of CO 2 at STP will be produced from the complete combustion of 23.2 g C 4 H 10 ? How many liters of CO 2 at STP will be produced from the complete combustion of 23.2 g C 4 H 10 ? What volume of oxygen will be required? What volume of oxygen will be required?

37 Gases and Reactions A few more details

38 Example How How many liters of CH 4 CH 4 at STP are required to completely react with 17.5 L of O 2 O 2 ? CH 4 CH 4 + 2O 2 2O 2  CO 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

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

40 Example How many liters of CO 2 at STP are produced by completely burning 17.5 L of CH 4 ? 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 CH 4 + 2O 2  CO 2 + 2H 2 O 17.5 L CH 4 1 L CH 4 2 L CO 2 = 35.0 L CH 4

41 Limiting Reagent 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. 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. The limiting reagent is the reactant you run out of first. The limiting reagent is the reactant you run out of first. The excess reagent is the one you have left over. The excess reagent is the one you have left over. The limiting reagent determines how much product you can make The limiting reagent determines how much product you can make

42 How do you find out? Do two stoichiometry problems. Do two stoichiometry problems. The one that makes the least product is the limiting reagent. The one that makes the least product is the limiting reagent. For example For example 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? 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?

43 If 10.6 g of copper reacts with 3.83 g S. How many grams of product will be formed? 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 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

44 Your turn If 10.1 g of magnesium and 2.87 L of HCl gas are reacted, how many liters of gas will be produced? If 10.1 g of magnesium and 2.87 L of HCl gas are reacted, how many liters of gas will be produced? How many grams of solid? How many grams of solid? How much excess reagent remains? How much excess reagent remains?

45 Your Turn II If 10.3 g of aluminum are reacted with 51.7 g of CuSO 4 how much copper will be produced? If 10.3 g of aluminum are reacted with 51.7 g of CuSO 4 how much copper will be produced? How much excess reagent will remain? How much excess reagent will remain?

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

48 Example 6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate 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 2Al + 3 CuSO 4  Al 2 (SO 4 ) 3 + 3Cu What is the actual yield? What is the actual yield? What is the theoretical yield? What is the theoretical yield? What is the percent yield? What is the percent yield?

49 Details Percent yield tells us how “efficient” a reaction is. Percent yield tells us how “efficient” a reaction is. Percent yield can not be bigger than 100 %. Percent yield can not be bigger than 100 %.

50 Homework!!


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