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Chemical Quantities – Ch. 9 Stoichiometry u Greek for “measuring elements” u The calculations of quantities in chemical reactions based on a balanced.

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Presentation on theme: "Chemical Quantities – Ch. 9 Stoichiometry u Greek for “measuring elements” u The calculations of quantities in chemical reactions based on a balanced."— Presentation transcript:

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2 Chemical Quantities – Ch. 9

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 2Na + 2H 2 O  2NaOH + H 2

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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 10 23 ) molecules of hydrogen and 1 x (6.02 x 10 23 ) molecules of oxygen form 2 x (6.02 x 10 23 ) 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 2 2.02 g H 2 1 moles H 2 =4.04 g H 2 1 moles O 2 32.00 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 18.02 g H 2 O 1 mole H 2 O = 36.04 g H 2 O 2H 2 + O 2   2H 2 O 36.04 g (H 2 + O 2 ) =36.04 g H 2 O

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

12 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

13 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 2 3.34 moles Al 2 O 3 2 moles Al 2 O 3 3 mole O 2 =5.01 moles O 2

14 Your Turn  2C 2 H 2 + 5 O 2  4CO 2 + 2 H 2 O u If 3.84 moles of C 2 H 2 are burned, how many moles of O 2 are needed? u How many moles of C 2 H 2 are needed to produce 8.95 mole of H 2 O? u If 2.47 moles of C 2 H 2 are burned, how many moles of CO 2 are formed?

15 Mole to Mole Conversions  2C 2 H 2 + 5 O 2  4CO 2 + 2 H 2 O u How many moles of C 2 H 2 are needed to produce 8.95 mole of H 2 O? u If 2.47 moles of C 2 H 2 are burned, how many moles of CO 2 are formed?

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

17 We can’t measure moles!! u What can we do? u We can convert grams to moles. u Periodic Table u Then use moles to change chemicals u Balanced equation u Then turn the moles back to grams. u Periodic table

18 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

19 Conversions  2C 2 H 2 + 5 O 2  4CO 2 + 2 H 2 O u How many moles of C 2 H 2 are needed to produce 8.95 g of 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 55.85 g Fe 1 mol Fe 2 mol Fe 3 mol Cu 1 mol Cu 63.55 g Cu =17.3 g Cu

21 2Fe + 3CuSO 4  Fe 2 (SO 4 ) 3 + 3Cu 0.181 mol Fe 2 mol Fe 3 mol Cu = 0.272 mol Cu 1 mol Cu 63.55 g Cu =17.3 g Cu

22 Could have done it 10.1 g Fe 55.85 g Fe 1 mol Fe 2 mol Fe 3 mol Cu 1 mol Cu 63.55 g Cu =17.3 g Cu

23 More Examples 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? u 3.74 mol of Mg would make how many moles of Si? u How many grams of MgCl 2 are produced along with 9.3 g of silicon?

24 For Example u The U. S. Space Shuttle boosters use this reaction  3 Al(s) + 3 NH 4 ClO 4  Al 2 O 3 + AlCl 3 + 3 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 ?

25 How do you get good at this?

26 Gases and Reactions

27 We can also change u Liters of a gas to moles u At STP u 0ºC and 1 atmosphere pressure u At STP 22.4 L of a gas = 1 mole u If 6.45 moles of water are decomposed, how many liters of oxygen will be produced at STP?

28 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 2 6.45 g H 2 O 18.02 g H 2 O 1 mol H 2 O 2 mol H 2 O 1 mol O 2 22.4 L O 2

29 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 ? u What volume of oxygen will be required?

30 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 O2O2 22.4 L O2O2 1 mol O2O2 2 O2O2 1 CH 4 1 mol CH 4 22.4 L CH 4 = 8.75 L CH 4 22.4 L O 2 1 mol O 2 1 mol CH 4 22.4 L CH 4

31 Avagadro 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.

32 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

33 Particles u We can also change between particles and moles. u 6.02 x 10 23 Molecules Atoms Formula units

34 Limiting Reactants u Available Ingredients 4 slices of bread 1 jar of peanut butter 1/2 jar of jelly u Limiting Reactant bread u Excess Reactants peanut butter and jelly

35 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

36 Limiting Reactants u Limiting Reactant used up in a reaction determines the amount of product u Excess Reactant added to ensure that the other reactant is completely used up cheaper & easier to recycle

37 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?

38 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 159.16 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 159.16 g Cu 2 S = 19.0 g Cu 2 S = 13.3 g Cu 2 S Cu is Limiting Reagent

39 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

40 Your turn  Mg(s) +2 HCl(g)  MgCl 2 (s) +H 2 (g) u If 10.1 mol of magnesium and 4.87 mol of HCl gas are reacted, how many moles of gas will be produced? u How much excess reagent remains?

41 Your Turn II 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?

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43 Percent Yield calculated on paper measured in lab

44 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

45 Percent Yield u When 45.8 g of K 2 CO 3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. K 2 CO 3 + 2HCl  2KCl + H 2 O + CO 2 45.8 g ? g actual: 46.3 g

46 Percent Yield 45.8 g K 2 CO 3 1 mol K 2 CO 3 138.21 g K 2 CO 3 = 49.4 g KCl 2 mol KCl 1 mol K 2 CO 3 74.55 g KCl 1 mol KCl K 2 CO 3 + 2HCl  2KCl + H 2 O + CO 2 45.8 g ? g actual: 46.3 g Theoretical Yield:

47 Percent Yield Theoretical Yield = 49.4 g KCl % Yield = 46.3 g 49.4 g  100 = 93.7% K 2 CO 3 + 2HCl  2KCl + H 2 O + CO 2 45.8 g 49.4 g actual: 46.3 g

48 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?

49 Details u Percent yield tells us how “efficient” a reaction is. u Percent yield can not be bigger than 100 %. u How would you get good at this?


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