2 Stoichiometry Greek for “measuring elements” The calculations of quantities in chemical reactions based on a balanced equation.We can interpret balanced chemical equations several ways.
3 In terms of Particles Element- atoms Molecular compound (non- metals)- moleculeIonic Compounds (Metal and non-metal) - formula unit
4 2H2 + O2 ® 2H2OTwo molecules of hydrogen and one molecule of oxygen form two molecules of water.2 Al2O3 ® 4Al + 3O22formula unitsAl2O3form4atomsAland3moleculesO22Na + 2H2O ® 2NaOH + H2
5 Look at it differently 2H2 + O2 ® 2H2O 2 dozen molecules of hydrogen and 1 dozen molecules of oxygen form 2 dozen molecules of water.2 x (6.02 x 1023) molecules of hydrogen and 1 x (6.02 x 1023) molecules of oxygen form 2 x (6.02 x 1023) molecules of water.2 moles of hydrogen and 1 mole of oxygen form 2 moles of water.
6 In terms of Moles 2 Al2O3 ® 4Al + 3O2 2Na + 2H2O ® 2NaOH + H2 The coefficients tell us how many moles of each kind
7 In terms of massThe law of conservation of mass appliesWe can check using moles2H2 + O2 ® 2H2O2.02 g H22 moles H24.04 g H2=1 moles H232.00 g O2=1 moles O232.00 g O21 moles O236.04 g H236.04 g Reactants
8 No mass has been created or destroyed Therefore…2H2 + O2 ® 2H2O18.02 g H2O36.04 g H2O2 moles H2O=1 mole H2O2H2 + O2 ® 2H2O36.04 g H2 + O2=36.04 g H2ONo mass has been created or destroyed
9 Your turnShow that the following equation follows the Law of conservation of mass.2 Al2O3 ® 4Al + 3O2
10 Mole to mole conversions 2 Al2O3 ® 4Al + 3O2every time we use 2 moles of Al2O3 we make 3 moles of O22 moles Al2O33 mole O2or3 mole O22 moles Al2O3
11 Mole to Mole conversions How many moles of O2 are produced when 3.34 moles of Al2O3 decompose?2 Al2O3 ® 4Al + 3O23.34 moles Al2O33 mole O2=5.01 moles O22 moles Al2O3
12 Your Turn2C2H2 + 5 O2 ® 4CO2 + 2 H2OIf 3.84 moles of C2H2 are burned, how many moles of O2 are needed?How many moles of C2H2 are needed to produce 8.95 mole of H2O?If 2.47 moles of C2H2 are burned, how many moles of CO2 are formed?
16 Given the following equation: 2 KClO3 2KCl + 3 O2How many moles of O2 can be produced by letting moles of KClO3 react?
17 2 KClO3 2KCl + 3 O2The strategy will be to convert to Remember, conversion factors are what you over what youmoles of KClO3moles of O2wanthave
18 KClO3 2KCl + O22312.00 moles of KClO3moles of O2moles of KClO3= 12 x (3/2)= 18
19 If ammonia, NH3, is burned in air, the following reaction takes place: 4NH3 + 3O2 2 N2 + 6H2OGiven that you started with 51.0 g of NH3, how many grams of water will be produced?
20 4NH3 + 3O2 2 N2 + 6H2OThe 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.gramsto molesmolesto molesmolesto grams
21 4NH3 + 3O2 2 N2 + 6H2O = 81 g of H2O 51.0 g NH3 grams NH moles NH moles H2O grams of H2O51.0 g NH31 mole NH318 g H2O6 mole H2O4 mole NH31 mole H2O17 g NH3= 81 g of H2O
22 20.0 g of silver(I)nitrate is reacted with an excess of sodium choride to produce silver(I) chloride AgNO3 + NaCl => AgCl + NaNO3What mass of silver(I) chloride is produced?
23 AgNO3 + NaCl => AgCl + NaNO3 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.gramsto molesmolesto molesmolesto grams
26 Mass in Chemical Reactions How much do you make?How much do you need?
27 We can’t measure moles!! What can we do? We can convert grams to moles.Periodic TableThen do the math with the moles.Balanced equationThen turn the moles back to grams.Periodic table
28 For example...If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how much solid copper would form?Fe + CuSO4 ® Fe2(SO4)3 + Cu2Fe + 3CuSO4 ® Fe2(SO4)3 + 3Cu1 mol Fe10.1 g Fe0.181 mol Fe=55.85 g Fe
29 2Fe + 3CuSO4 ® Fe2(SO4)3 + 3Cu3 mol Cu0.272 mol Cu0.181 mol Fe=2 mol Fe63.55 g Cu0.272 mol Cu=17.3 g Cu1 mol Cu
30 Could have done it 1 mol Fe 63.55 g Cu 10.1 g Fe 3 mol Cu 55.85 g Fe =17.3 g Cu
31 More ExamplesTo make silicon for computer chips they use this reactionSiCl4 + 2Mg ® 2MgCl2 + SiHow many grams of Mg are needed to make 9.3 g of Si?How many grams of SiCl4 are needed to make 9.3 g of Si?How many grams of MgCl2 are produced along with 9.3 g of silicon?
32 For Example The U. S. Space Shuttle boosters use this reaction 3 Al(s) + 3 NH4ClO4 ® Al2O3 + AlCl3 + 3 NO + 6H2OHow much Al must be used to react with 652 g of NH4ClO4 ?How much water is produced?How much AlCl3?
33 We can also change Liters of a gas to moles At STP 25ºC and 1 atmosphere pressureAt STP 22.4 L of a gas = 1 moleIf 6.45 moles of water are decomposed, how many liters of oxygen will be produced at STP?
34 For ExampleIf 6.45 grams of water are decomposed, how many liters of oxygen will be produced at STP?H2O ® H2 + O21 mol H2O1 mol O222.4 L O26.45 g H2O18.02 g H2O2 mol H2O1 mol O2
35 Your TurnHow many liters of CO2 at STP will be produced from the complete combustion of 23.2 g C4H10 ?What volume of oxygen will be required?
37 ExampleHow many liters of CH4 at STP are required to completely react with 17.5 L of O2 ?CH4 + 2O2 ® CO2 + 2H2O22.4 L O21 mol O21 mol CH422.4 L CH41 mol O21 mol CH422.4 L CH417.5 L O222.4 L O22 mol O21 mol CH4= 8.75 L CH4
38 Avagadro told usEqual volumes of gas, at the same temperature and pressure contain the same number of particles.Moles are numbers of particlesYou can treat reactions as if they happen liters at a time, as long as you keep the temperature and pressure the same.
39 ExampleHow many liters of CO2 at STP are produced by completely burning L of CH4 ?CH4 + 2O2 ® CO2 + 2H2O2 L CO217.5 L CH4= 35.0 L CH41 L CH4
40 Limiting ReagentIf 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 excess reagent is the one you have left over.The limiting reagent determines how much product you can make
41 How do you find out? Do two stoichiometry problems. The one that makes the least product is the limiting reagent.For exampleCopper 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?
42 If 10. 6 g of copper reacts with 3. 83 g S If 10.6 g of copper reacts with 3.83 g S. How many grams of product will be formed?2Cu + S ® Cu2SCu is Limiting Reagent1 mol Cu1 mol Cu2Sg Cu2S10.6 g Cu63.55g Cu2 mol Cu1 mol Cu2S= 13.3 g Cu2S= 13.3 g Cu2S1 mol S1 mol Cu2Sg Cu2S3.83 g S32.06g S1 mol S1 mol Cu2S= 19.0 g Cu2S
43 Your turnIf 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 much excess reagent remains?
44 Your Turn IIIf 10.3 g of aluminum are reacted with 51.7 g of CuSO4 how much copper will be produced?How much excess reagent will remain?
46 Yield The amount of product made in a chemical reaction. There are three typesActual yield- what you get in the lab when the chemicals are mixedTheoretical yield- what the balanced equation tells you you should make.Percent yield = Actual x 100 % Theoretical
47 Example6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate.2Al + 3 CuSO4 ® Al2(SO4)3 + 3CuWhat is the actual yield?What is the theoretical yield?What is the percent yield?
48 Details Percent yield tells us how “efficient” a reaction is. Percent yield can not be bigger than 100 %.