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Transcript
S Stoichiometry STOICHIOMETRY S Quantitative measurements are those that involve actual number relationships, relationships, not just human judgment. Ex. 3.21 cm, 5.42 moles, 342.65 g, 24.1 L not just, “It was really, really a lot” 1 STOICHIOMETRY S We will look at quantitative measurements in chemical reactions. Mass and mole relationships in chemical reactions Limiting Reactants in a reaction Theoretical and percent yield Solution concentrations and preparations (acid/base reactions) S Hamburger Analogy Recipe for a bacon double cheeseburger is: 1 hamburger bun 2 hamburger patties 2 slices of cheese 4 strips of bacon 2 S If I have five bacon double cheeseburgers: •How many hamburger buns do I have? •How many hamburger patties do I have? •How many slices of cheese do I have? •How many strips of bacon do I have? S How many bacon double cheeseburgers can you make if you start with: a) 1 bun, bun, 2 patties patties,, 2 slices cheese, cheese, 4 strips bacon b) 2 bun, bun, 4 patties patties,, 4 slices cheese, cheese, 8 strips bacon c) 1 dozen bun, bun, 2 dz. patties patties,, 2 dz. slices cheese, cheese, 4 dozen strips bacon d) 1 mole bun, bun, 2 mole patties patties,, 2 mole slices cheese, cheese, 4 mole strips bacon e) 10 bun, bun, 20 patties patties,, 2 slices cheese, cheese, 40 strips bacon 3 S 10 bun, bun, 20 patties patties,, 2 slices cheese, cheese, 40 strips bacon 1 BacDbl Cheeseburger If you had fixings for 100 bacon double cheeseburgers, but when you were cooking you burned 20 paddies. What percentage of the bacon double cheeseburgers do you actually make? 90% BacDbl Cheeseburger At this point what is the limiting ingredient? Why? DCBURGERS AND CHEMISTRY S How many sandwiches can you make? 12 ____ slices of bread 10 + ____ burger patties 5 = ____ burgers What is left over? 2 slices of bread What is the limiting ingredient? burger 4 EQUATION REVIEW C2H5OH + 3O2 → 2CO2 + 3H2O reactants products EQUATION REVIEW C2H5OH + 3O2 → 2CO2 + 3H2O reactants When the equation is balanced it has quantitative significance: 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water 5 BURGERS AND CHEMISTRY S The math and the concepts are identical to the Previous slide. The only difference between burger and formula is the names in the recipe. Here are two examples of chemical recipes: 1. 2. Na+ + SO42- → Na2SO4 1 mole of H2SO4 + 2 mole NaOH → 1 mole Na2SO4 + 2 mole H2O PROPORTIONAL RELATIONSHIPS S Stoichiometry mass relationships between substances in a chemical reaction based on the mole ratio Mole Ratio indicated by coefficients in a balanced equation 2 Mg2+ + O22- → 2 MgO 6 S 2Na+ + Cl2- → 2NaCl If I have 1 mole of NaCl How many moles of sodium created it? How many moles of diatomic chloride created it? 1 0.5 1 mole of H2SO4 + 2 mole NaOH → 1 mole Na2SO4 + 2 mole H2O If I want to make 5 moles of Na2SO4: How many moles of H2SO4 do I need? How many moles of NaOH do I need? MOLE RATIO CONVERSIONS C2H5OH + 3O2 → 2CO2 + 3H2O 1 2 S How many moles of CO2 can I make if I have: 12 mole of C2H5OH 2moleCO2 24 mole of CO2 1moleC2 H 5OH .90 mole of O2 2moleCO2 3moleO2 0.6 mole of CO2 How many moles of C2H5OH did I need to make 1moleC2 H 5OH 10 mole of H2O 3.3 mole of C2H5OH 3moleH 2O 0.65 mole of CO2 1moleC2 H 5OH 2moleCO2 3.3 mole of C2H5OH 7 MOLE RATIO CONVERSIONS S 1 mole of H2SO4 + 2 mole NaOH → 1 mole Na2SO4 + 2 mole H2O How many moles of Na2SO4 can I make if I have: 1 mole of H2SO4 1moleNa2 SO4 1 mole of Na2SO4 1moleH 2 SO4 1moleNa2 SO4 0.1 mole of Na SO 2 4 2moleNaOH How many moles of H2SO4 did I need to make 17 mole of H2O 1moleH 2 SO4 8.5 mole of Na2SO4 2moleH 2O 0.65 mole of Na2SO4 0.65 mole of Na2SO4 1moleH 2 SO4 .20 mole of NaOH 1moleNa2 SO4 S 1 mole of H2SO4 + 2 mole NaOH → 1 mole Na2SO4 + 2 mole H2O How many moles of Na2SO4 can I make if I have: 1 mole of H2SO4 and 2 moles of NaOH 10 mole of H2SO4 and 20 mole of NaOH 0.1 mole of H2SO4 and 0.2 mole of NaOH 1 mole of H2SO4 and 20 mole of NaOH 0.42 mole of H2SO4 and 0.65 mole of NaOH 1 10 .1 1 .325 8 The key to understanding… The coefficients in a balanced equation give the relative amounts (in moles) of reactants and products. The coefficients represent moles… 2NaCl → 2Na + Cl2 S S There are ____ moles of salt. There are ____ moles of sodium. There is ____ mole of diatomic Cl. 9 MOLAR MASS S A substance’s molar mass (molecular weight) is the mass in grams of one mole of the compound. CO2 = 44.01 grams per mole H2O = 18.02 grams per mole Ca(OH)2 = 74.10 grams per mole STOICHIOMETRY 10 STEPS TO STOICHIOMETRY STEP 1 S S Write the balanced equation! 11 STEP 2 S Find the number of moles of the given substance STEP 3 S Use the balanced equation to find the mole ratio of given substance to moles of unknown 12 S STEP 4 Convert moles of unknown to whatever quantity is specified STOICHIOMETRY N2 + 3 H2 → S 2NH3 In a balanced chemical equation, the coefficients allow us to calculate how much of a substance is used, produced, or needed in the reaction. 13 STOICHIOMETRY N2(g) + 3H2 (g) → S 2NH3 From this equation we can determine that 1 mole of Nitrogen gas (28.02 g ) reacts with 3 moles of Hydrogen gas (6.06 g) to produce 2 moles of ammonia (28.02g + 6.06g=34.08g of ammonia) Always remember the conservation of matter STOICHIOMETRY RECAP S How many moles of NaOH will react with 3.21 moles of sulfuric acid? Step 1. Balanced the chemical equation. Step 2. Write down the given numeric info. If it is not in mole units convert it to moles using the molecular mass as a conversion factor. Step 3. Now convert from the moles of starting substance to the moles of the desired substance by using a mole ratio from the chemical reaction. reaction. Step 4. If the problem is to find grams of the substance, convert the moles to grams using the molecular mass conversion factor. If the problem is to find atoms of the substance, convert moles to atoms. 14 STOICHIOMETRY S How many moles of NaOH will react with 3.21 moles of sulfuric acid (H2SO4)? Now lets work out the example problem: Step 1 H2SO4 + 2NaOH → Na 2SO4 + 2H2O Step 2 (Given in moles; so is not needed) Step 3 (Note that step 2 is not needed) 3.21 moles H2SO4 x 2 moles NaOH 1 mol H2SO4 = 6.42 moles NaOH Step 4 (Asked for moles; so is not needed) MOLE RATIOS S These mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemical Example: How many moles of chlorine is needed to react with 5 moles of sodium (without any sodium left over)? 2 Na + Cl2 → 2 NaCl 5 moles Na 1 mol Cl2 = 2.5 moles 2 mol Na Cl2 15 MOLE RATIO - THE BRIDGE S EVERYTIME YOU GO ACROSS A BRIDGE I EXPECT YOU TO THINK ABOUT ME IN THIS CLASSROOM STILL TEACHING CHEMISTRY AS YOU DRIVE/WALK ACROSS. IF YOU TAKE PHYSICS WITH ME YOU WILL GET TO BUILD BRIDGES. THE BRIDGE IS THE MOST IMPORTANT THING I CAN TEACH YOU TO UNDERSTANDING CHEMICAL CALCULATIONS. MOLE RATIO S Shows the molemole-to to--mole ratio between two of the substances in a balanced equation Derived from the coefficients of any two substances in the equation 16 S MOLE RATIO → 3 H2(g) + N2(g) 2 NH3(g) A. A mol ratio for H2 and N2 1) 3 mol N2 2) 1 mol N2 1 mol H2 3) 1 mol N2 3 mol H2 2 mol H2 B. A mol ratio for NH3 and H2 is 1) 1 mol H2 2) 2 mol NH3 2 mol NH3 3) 3 mol N2 3 mol H2 2 mol NH3 S MOLE RATIO → 4 Fe + 3 O2 2 Fe2O3 What is the mole ratio of the compounds? Fe and O2 4 mol Fe 3 mol O2 or 3 mol O2 4 mol Fe Fe and Fe2O3 4 mol Fe 2 mol Fe2O3 or 2 mol Fe2O3 4 mol Fe 17 S MOLE RATIO 4 Fe + 3 O2 → 2 Fe2O3 O2 and Fe2O3 3 mol O2 2 mol Fe2O3 and 2 mol Fe2O3 3 mol O2 Which MR we use depends on what numbers we Know and which we are trying to get rid of. Mole--Mole Conversions Mole S How many moles of sodium chloride will be produced if you react 2.6 moles of chlorine gas with an excess (more than you need) of sodium metal? 2 Na + Cl2 → 2 NaCl 18 S MOLE TO MASS & MASS TO MOLE MASS--MASS Conversions S MASS Most of the time in chemistry, the amounts are given in grams instead of moles We still go through moles and use the mole ratio, but now we also use molar mass to get to grams Example: How many grams of chlorine are required to react completely with 5.00 moles of sodium to produce sodium chloride? 2 Na + Cl2 → 2 NaCl 5.00 moles Na 1 mol Cl2 2 mol Na 70.90 g Cl2 1 mol Cl2 = 177g Cl2 19 MOLE--MASS MOLE S How many moles of Fe2O3 are produced when 6.0 moles O2 react with Iron? → 4 Fe + 3 O2 2 Fe2O3 How many grams of Fe2O3 would that be? MOLE--MASS MOLE 4 Fe + 6.0 mol O2 x 2 3 3 O2 → S 2 Fe2O3 mol Fe2O3= 4.0 mol Fe2O3 mol O2 4.0 mol Fe2O3 x 159.6 g Fe2O3 = 640 g Fe2O3 mol Fe2O3 20 MASS--MASS MASS S How many grams of Fe are needed to react with 12.0 mol of O2? 4 Fe + 3 O2 → 2 Fe2O3 How many grams of Fe are needed? MASS--MASS MASS S How many grams of Fe are needed to react with 48g of O2? Fe + O2 → Fe2O3 21 MOLE--MASS MOLE 4 Fe + 3 O2 → S 2 Fe2O3 How many grams of O2 are needed to produce 0.400 mol of Fe2O3? 0.400 mol Fe2O3 3 mol O2 2 mol Fe2O3 32.0 g O2 1 mol O2 = 19.2 g O2 MASS TO MASS S 22 Mass--Mass Conversions Mass S Usually we are given a starting mass and want to find out the mass of a product. This calculated mass is called theoretical yield. Theoretical yield is how much of another reactant we need to completely react with it (no leftover ingredients!) Now we go from grams to moles, mole ratio, and back to grams of compound. Many have trouble at conversions. So I have a visual model to help us, it is called the horseshoe. Mass--Mass Conversions Mass aX + bY → cZ S 1. Balance an equation in moles 2. Convert given grams/atoms X to moles X HERE IS WHAT THE CONVERSION 3. Use mole factor to give desired moles Y/Z LOOKS 4. Convert moles Y/Z toLIKE grams/atoms Y/Z atoms/grams (given) X atoms/grams (desired) Y/Z X MOLAR MASS / MOLAR MASS or Avogadro's # moles (given) X MOLE RATIO moles (desired) Y/Z atoms/g X x 1 mol X x b/c mol Y/Z x atoms/grams Y/Z atoms / mm X a mol X 1 mol Y/Z 23 HORSESHOE ANALOGY atoms/grams (given) (X) Molar Mass S atoms/grams (desired) The horseshoe pictorially shows the stoichiometric process. The only thing you will really mess up is the MOLE RATIO. RATIO. Make to use (/) Molar Mass Want / Have. Have. moles (given) moles (desired) mole ratio moles (desired) moles (given) “HOW TO” HORSESHOE Grams N2 How many moles of Hydrogen are required to react with 103.25 g of Nitrogen (AKA mole ratio) in the Haber Process to produce ammonia? (/) Molar Mass N2 S Grams H2 (X) Molar Mass H2 moles N2 moles H2 mole ratio moles H2 moles N2 24 MASS--MOLE MASS S How many moles of Hydrogen are required to react with 103.25 g of Nitrogen (AKA mole ratio) in the Haber Process to produce ammonia? H2 + N2 → NH3 Step 1. Balance 3H2 + N2 → 2NH3 Step 2. Determine Mole of Known 103.25 g N2 x 1mole N2 = 3.68 moles of N2 28.02 g N2 Step 3. Multiply Mole Ratio by Known Moles 3.68 moles N2 x 3 moles H2 = 11.4mole H2 1 mol N2 MASS--MOLE MASS S How many moles of Hydrogen are required to react with 103.25 g of Nitrogen (AKA mole ratio) in the Haber Process to produce ammonia? 3 H2 + N2 → 2 NH3 103.25 g N2 1mole N2 3 moles H2 =11.4mol H2 28.02 g N2 1 mol N2 25 MASS--MASS MASS S The reaction between H2 and O2 produces 13.1 g of water.. How many grams of O2 reacted? water H2 (g) + O2 (g) → 2 H2O (g) HORSESHOE Grams H2O The reaction between H2 and O2 produces 13.1 g of water. How many grams of O2 reacted? (/) Molar Mass H2O S Grams O2 (X) Molar Mass O2 moles H2O moles O2 mole ratio moles O2 moles H2O 26 MASS TO MASS S The reaction between H2 and O2 produces 13.1 g of water.. How many grams of O2 reacted? water 2 H2 + O2 → 2 H2O 13.1 g H2O x 1 mole H2O x 1 mol O2 x 32.0 g O2 = _11.6_ g O 2 18.0 g H2O 2 mol H2O 1 mol O2 HORSESHOE Grams CO2 2 S Acetylene gas C2H2 burns in the oxyactylene torch for welding. How many grams of C2H2 are Grams C2H2 burned if the reaction produces 75.0 g of CO2? C2H2+ 5O2 →4 CO2 + 2 H2O (/) Molar Mass CO2 (X) Molar Mass C2H2 moles CO2 moles C2H2 mole ratio moles C2H2 moles CO2 27 S MASS TO MASS Acetylene gas C2H2 burns in the oxyactylene torch for welding. How many grams of C2H2 are 2 4 75.0 g 2 burned if the 5 reaction produces of CO2? C2 H2 + O2 → CO2 + H2O 75.0 g CO2 x 1 mol CO2 x 2 mol C2H2 x 26.0 g C2H2 44.0 g CO2 4 mol CO2 1 mol C2H2 = 22.2 g C2H2 HORSESHOE Grams N2 Calculate how many grams of ammonia are produced when you react 2.00g of nitrogen with excess hydrogen. S Grams NH3 N2 + 3 H2 → 2 NH3 (/) Molar Mass N2 (X) Molar Mass NH3 moles N2 Moles NH3 mole ratio moles NH3 moles N2 28 S MASS--MASS MASS Calculate how many grams of ammonia are produced when you react 2.00g of nitrogen with excess hydrogen. N2 + 3 H2 → 2 NH3 2.00g N2 1 mol N2 2 mol NH3 28.02g N2 1 mol N2 17.06g NH3 1 mol NH3 = 2.4 g NH3 HORSESHOE Grams H2O Calculate how many grams of Oxygen produces 13.1 g H2O in the following reaction. S Grams O2 2H2 + O2 → CO2 + 2H2O (/) Molar Mass H2O (X) Molar Mass O2 moles H2O Moles O2 mole ratio moles O2 moles H2O 29 MASS--MASS MASS S Calculate how many grams of Oxygen produces 13.1 g H2O in the reaction. → 2H2 + O2 CO2 + 13.1 g H2O x 1 mol H2O x 1 mol O2 x 18.0 g H2O 2 mol H2O = 2H2O 32.0 g O2 1 mol O2 11.6 g O2 HORSESHOE Grams H2 How many grams of Hydrogen are required to react with 100.00g of Oxygen in the production of water? (/) Molar Mass H2 S Grams O2 (X) Molar Mass O2 moles H2 moles O2 mole ratio moles O2 moles H2 30 MASS TO MASS S How many grams of Hydrogen are required to react with 100.00g of Oxygen in the production of water? 2 H2 + O2 → 2 H2O 100.0 g O2 x 1 mole O2 x 2mol H2 x 2.02 g H2 = 12.60 g H2 32.0 g O2 1 mol O2 1 mol H2 MASS--MASS MASS S If you start with 58.44g of salt (NaCl (NaCl), ), how many grams of Cl2 are produced when salt is decomposed? 2NaCl → 2Na + Cl2 58.44g NaCl 1 mol NaCl ____ gNaCl = 1 mol Cl2 70.90 g Cl2 2 mol NaCl 1 mol Cl2 Cl2 31 MASS--MASS MASS Iron combines with Sulfur according to the following reaction: Fe (s) + S(l) → S FeS (s) In an experiment, .25 mol FeS (s) is created in a reaction of iron and excess S(l) . Calculate the mass of Fe that was used. HORSESHOE Grams FeS In an experiment, .25 mol FeS (s) is created in a reaction of iron and excess S(l) . (/) Molar Mass FeS S Grams Fe (X) Molar Mass Fe moles FeS moles Fe mole ratio moles Fe moles FeS 32 MASS--MASS MASS 0.25 mol FeS x 1 mol Fe x 55.85g Fe 1 mol FeS 1 mol Fe S = 14g FeS “HOW TO” HORSESHOE Grams N2 How many moles of Hydrogen are required to react with 103.25 g of Nitrogen (AKA mole ratio) in the Haber Process to produce ammonia? (/) Molar Mass N2 S Grams H2 (X) Molar Mass H2 moles N2 moles H2 mole ratio moles H2 moles N2 33 MASS--MASS MASS S N2 + 3 H2 → 2 NH3 2.00g N2 1 mol N2 2 mol NH3 17.06g NH3 28.02g N2 1 mol N2 1 mol NH3 = 2.4 g NH3 HORSESHOE Grams H2O Calculate how many grams of Oxygen produces 13.1 g H2O in the following reaction. (/) Molar Mass Na moles H2O S Grams O2 (X) Molar Mass O2 moles O2 mole ratio moles O2 moles H2O 34 MASS--MASS MASS S Calculate how many grams of Oxygen produces 13.1 g H2O in the following reaction. 2H2 + O2 → CO2 + 2H2O 13.1 g H2O x 1 mol H2O x 1 mol O2 x 32.0 g O2 18.0 g H2O 2 mol H2O 1 mol O2 = 11.6 g O2 MASS--MASS MASS S We can also start with mass and convert to moles of product or another reactant We use molar mass and the mole ratio to get to moles of the compound of interest Calculate the number of moles of ethane (C2H6) needed to produce 10.0 g of water 2 C2H6 + 7 O2 → 4 CO2 + 6 H20 10.0 g H2O 1 mol H2O 2 mol C2H6 = 0.185 mol 18.0 g H2O 6 mol H20 C2H6 = 0.185 mol C2H6 x 30.0g/mol =5.5g C2H6 35 MASS--MASS MASS 4 Fe + 48 g O2 x → 3 O2 mol O2 = S 2 Fe2O3 1.5 mol O2 32.0g O2 1.5 mol O2 x 4mol Fe= 3 2.0 mol Fe mol O2 2.0 mol Fe x 55.85 g Fe = 1 mol Fe 111.4 g Fe MASS--MASS MASS 4 Fe + 3 O2 12.0 mol O2 x → S 2 Fe2O3 4 mol Fe = 3 mol O2 16.0 mol Fe x 55.85 g Fe = 1 mol Fe 16.0 mol Fe 894g Fe 36 MASS--MASS MASS S How many grams of Fe are needed to react with 68g of Fe2O3? 4 Fe → 2 Fe2O3 3 O2 + How many grams of Fe are needed? MASS--MASS MASS 4 Fe + 68 g Fe2O3 x 3 O2 → S 2 Fe2O3 mol Fe2O3 = .425 mol Fe2O3 160.0g Fe2O3 .425 mol Fe2O3 X 4 2 mol Fe = .85 mol Fe mol Fe2O3 .85 mol Fe x 55.85 g Fe = 1 mol Fe2O3 47.58 g Fe 37 S aX + bY → cZ mol X 3 mol Y/Z mol X = Y/Z 1 mol Y S MASS TO ATOM ATOM,MOLECULE,PARTICLE, CRIDER 38 •Example: Calculate how many molecules of Carbon Dioxide is produced with 18.0g H2O in the following reaction. S •Grams H2O •CH4 + 2O2 → CO2 + 2H 2H2O •Grams CO2 •(X) Molar •(/) Molar • Mass CO2 • Mass H2O • •moles•(X) AV # •Atoms •moles •atoms •CO2 • H2 O • H2 O • 1moles C O 2 •(/) AV # • CO2 • 2moles H2O S The reaction between H2 and O2 produces 3.0x1024 atoms of water. How many atoms of O2 reacted? 39 HORSESHOE atoms H2O S The reaction between H2 and O2 produces 3.0x1024 atoms of water. How many molecules of O2 reacted? (/) AV # (X) AV # Atoms O2 moles O2 moles H2O mole ratio moles O2 moles H2O MASS TO MASS S The reaction between H2 and O2 produces 3.0x1024 atoms of water. How many atoms of O2 reacted? 2 H2 + O2 3x1024 atoms H2O 1 mole H2O 6.022x1023 H2O → 2 H2O 1 mol O2 2 mol H2O 6.022x1023 O2 = 1 mol O2 1.5 x 1024 atoms O2 40 HORSESHOE Grams Na S How many O2 molecules will react with 505g of Na to form Na2O? (/) Molar Mass Na (X) AV # Atoms O2 moles O2 moles Na mole ratio moles O2 moles Na MASS TO ATOMS S How many O2 atoms will react with 505g of Na to form Na2O? 4 Na + O2 → 2 Na2O 505 g Na x 1 mol Na x 1 mol O2 x 6.02 x 1023 atoms O2 23.0 g Na 4 mol Na 1 mol O2 = 3.30 x 1024 atoms O2 41 HORSESHOE Grams Fe2O3 S How many atoms of Fe are needed to create 2000.g of Fe2O3? (/) Molar Mass Fe2O3 (X) AV # moles Fe moles Fe2O3 Atoms Fe mole ratio moles Fe moles Fe2O3 MOLE--MASS MOLE 4 Fe + 3 O2 → S 2 Fe2O3 2000. g Fe2O3 x mol Fe2O3 = 12.50 mol 160.0g Fe2O3 12.50 mol Fe2O3 X 4 2 mol Fe = 25.00 mol Fe mol Fe2O3 25 mol Fe x 6.022x1023 g Fe = 1.500x1025 Fe 1 mol Fe 42 HORSESHOE S How many molecules of Fe2O3 are needed to react with 525g of Fe Fe? ? Grams Fe (/) Molar Mass Fe (X) AV # moles Fe2O3 moles Fe Atoms Fe2O3 mole ratio moles Fe2O3 moles Fe ATOMS--MASS ATOMS 4 Fe + 525 g Fe x 3 O2 → S 2 Fe2O3 1 mol Fe = 9.4 mol Fe 55.85g Fe 9.4 mol Fe X 2 4 mol Fe2O3 = 4.7mol Fe2O3 mol Fe 4.7 mol Fe x 160.0 g Fe2O3 = 1 mol Fe2O3 752 g Fe2O3 4.7 mol Fe x 6.022x1023 g Fe = 2.80x1024 molecules Fe 1 mol Fe 43 HORSESHOE How many grams Na will react with 5.0 x 1023 of O2 to form Na2O? S Grams Na (X) Molar Mass Na (X) AV # Atoms O2 moles O2 moles O2 moles Na moles Na S MASS TO ATOMS How many grams Na will react with 5.0 x 1023 of O2 to form Na2O? 4 Na + O2 5.0x1023 O2 x 1 mol O2 6.022x1023 O2 → 2 Na2O x 4 mol Na x 23.0 g Na 1 mol O2 1 mol Na = 76 g Na 44 SOLUTION STOICHIOMETRY CONCENTRATION S S The measure of the amount of solute in a certain amount of solution Solute the lesser quantity which is dissolved in the substance of greater quantity ,what ,what is being dissolved Solvent is the substance in greater quantity, what is doing the Dissolving Solution the mixture of solvent and solute 45 Concentration S Dilute small amount of solute in the solution Concentrated large amount of solute in the solution These are vague terms without definite boundaries They can be used to compare solutions MOLARITY S Most often used to specify the concentration of a solution MOLARITY = number of moles of solute in one liter of solution M= moles of solute liters of solution 46 MOLARITY S What is the molarity of a solution that has 3.21 moles of HCl dissolved in 2.4 L of solution? Molarity = moles of solute = 3.21 moles HCl Liters of solution 2.4 L solution = 1.34 M MOLARITY S 3.7 moles of HCl is added to water to make 500. mL of solution. What is the molarity? Molarity = moles of solute = 3.7 moles HCl Liters of solution 0.500 L solution = 7.4 M 47 MOLARITY S How many liters of 0.20 M solution can be from 3.00 mol of NaCl? Molarity = moles of solute = 3.0 moles NaCl Molarity 0.20 mol/L = 15L MOLARITY S How many moles of NaCl are needed to make 11.0 L of 0.15 M solution? How many grams of NaCl would that be? Moles solute = (Liters) (Molarity) = (11.0L)( (11.0L)(0.20 0.20 mol/L) = 1.7 mol NaCl 1.7 Moles x = 58.45g 1 mol = 99. g NaCl 48 MOLARITY S How many grams of sodium sulfate (Na2SO4 ) are required to prepare a 250 mL mL,, 0.683 M solution? 250 mL x 1L = 0.250 L 1000mL 0.250 L x 0.683 moles Na 2 SO 1 liter solution 4 = 0.171 moles Na 2SO4 0.171 moles Na2SO4 x 142.04 g Na 2SO4 = 24.3 g 1 mole Na2SO4 Na2SO4 21.0 g of NaOH is dissolved in enough water to make 500. mL of solution. What is the molarity? S 1mol NaOH = 0.525 mol NaOH 39.9970g NaOH moles of solute 0.525 mol M= = = 1.05 M NaOH liters of solution 0.500 L 21.0gNaOH × 49 S •Example: Calculate how many molecules of Carbon Dioxide is produced with 13.1 g H2O in the following reaction. S •Grams H2O 2H2O •CH4 + 2O2 → CO2 + 2H •Grams CO2 •(X) Molar •(/) Molar • Mass CO2 • Mass H2O • •Atoms •moles•(X) AV # •moles •Molecules • H2 O •CO2 • H2 O • 1moles CO2 •(/) AV # • CO2 • 2moles H2O •(X) M H2O •Liters H2O •(/) M CO2 •Liters CO2 50 Calculate how many molecules of Carbon Dioxide is produced with 13.1 g H2O in the following reaction. CH3 + 2O2 → CO2 + 2H 2H2O Grams H2O Atoms H2O (/) Molar Mass H2O (X) Molar Mass CO2 moles H2O moles CO2 (/) AV # 1moles CO2 2moles H2O Grams CO2 (X) AV # Molecules CO2 (/) (X) M H2O Liters H2O S M CO2 Liters CO2 HORSESHOE S How many liters of 3.0 M NaOH will react with 500mL of 6.0 M sulfuric acid? H2SO4 + 2NaOH → Na 2SO4 + 2H2OH moles H2SO4 (X) M H2SO4 Liter H2SO4 mole ratio moles NaOH moles H2SO4 moles NaOH (/) M NaOH Liter NaOH 51 S How many liters of 3.0 M NaOH will react with 500mL of 6.0 M sulfuric acid? H2SO4 + 2NaOH → Na 2SO4 + 2H2OH .5L H2SO4 x 6.0 moles H2SO4 = 3.0 moles H2SO4 1 L H2SO4 3.0 moles H2SO4 x 2moles NaOH = 6.0 moles NaOH 1 mol H2SO4 6.0 moles NaOH x 1 Liters NaOH = 2.0 L NaOH 3.0 mol NaOH HORSESHOE S How many liters of 3.0 M Sulfuric Acid will react with 3.21 moles of Sodium Hydroxide ? How many molecules of Sulfuric Acid ? H2SO4 + 2NaOH → Na 2SO4 + 2H2OH (X) AV # moles NaOH moles H2SO4 moles NaOH moles H2SO4 Molecules H2SO4 (/) M H2SO4 Liter H2SO4 52 S How many liters of 3.0 M Sulfuric Acid will react with 3.21 moles of Sodium Hydroxide ? How many molecules of Sulfuric Acid ? H2SO4 + 2NaOH → Na 2SO4 + 2H2OH 3.21 moles NaOH x 1 moles H2SO4 = 1.61 moles H2SO4 2 mol NaOH 1.61 moles H2SO4 x 1 Liters H2SO4 = .537 Liters H2SO4 3.0 mol H2SO4 1.61 moles H2SO4 x 6.022x1023 atoms = 9.69 x1024 atoms 1.0 mol H2SO4 HORSESHOE grams Zn S Zinc reacts with acids to produce H2 gas. If 10g of Zn reacts with 2.5M HCl, how many Liters of HCl is needed to convert all Zn? (/) Molar Mass Zn moles Zn mole ratio moles HCl moles Zn moles HCl (/) M HCl L HCl 53 S Zinc reacts with acids to produce H2 gas. If 10g of Zn reacts with 2.5M HCl, how many Liters of HCl is needed to convert all Zn? Zn (s) + 2 HCl (aq aq)) 10.0 g Zn 1 mol Zn 65.39g Zn → ZnCl2 (aq aq)) + H2 (g) 2 mol HCl 1.00L 1 mol Zn 2.5 mol HCl = 0.122 L HCl HORSESHOE How many grams of silver will be formed from 1.2 mL of .5M g AgNO3 and excess copper ? S Grams Ag (X) Molar Mass Ag moles AgNO3 moles Ag moles AgNO3 moles Ag (X) Molarity AgNO3 Liters AgNO3 54 S How many grams of silver will be formed from 1.2 mL of .5M AgNO3 and excess copper ? Cu + 2AgNO3 → 2Ag + Cu(NO3)2 .5M 1.0 mol AgNO3 AgNO 3 2.0 mol Ag 107.87 g Ag .0012 L AgNO3 2 mol AgNO3 1 mol Ag = .06g Ag HORSESHOE S How many grams of Cu are required to react with 1.5 L of 0.10M AgNO3? Grams Cu (X) Molar Mass Cu moles AgNO3 moles Cu moles AgNO3 moles Cu (X) Molarity AgNO3 Liters AgNO3 55 How many grams of Cu are required to react with 1.5 L of 0.10M AgNO3? S Cu + 2AgNO3 → 2Ag + Cu(NO3)2 1.5 L .10 mol AgNO3 1 mol Cu 63.55 g Cu 1L 2 mol AgNO3 1 mol = 4.8 g Cu Cu S LIMITING REACTANTS 56 •Example: Calculate how many molecules of Carbon Dioxide is produced with 13.1 g H2O in the following reaction. S •Grams H2O •CH4 + 2O2 → CO2 + 2H 2H2O •Grams CO2 •(X) Molar •(/) Molar • Mass CO2 • Mass H2O • •moles•(X) AV # •Atoms •moles •Molecules •CO2 • H2 O • H2 O • 1moles C O 2 •(/) AV # • CO2 • 2moles H2O •(X) M H2O •Liters H2O •(/) M CO2 •Liters CO2 LIMITING REACTANTS S The limiting reactant is the reactant present in the smallest stoichiometric amount Which of these will limit the reaction? Not Necessarily. Why? 57 LIMITING REACTANTS S Most of the time in chemistry we have more of one reactant than we need to completely use up other reactant. That reactant is said to be in excess (there is too much). The other reactant limits how much product we get. Once it runs out, the reaction s. This is called the limiting reactant. reactant. LIMITING REACTANTS S In the example below, the O2 would be the excess Reactant 58 LIMITING REACTANTS S We have to calculate how much (in moles) of a product we can get from each of the reactants to determine which reactant is the limiting one. The lower amount of moles of a product is the correct answer. The reactant that makes the least amount (in moles) of product is the limiting reactant. reactant. Once you determine the limiting reactant, you should ALWAYS start with it! LIMITING REACTANTS S N2 + O2 → 2NO If you react 1 mole of N2 with 1 mole of O2 you will get 2 moles of NO What happens if you start with 2 moles of O2 and 6 moles of N2? 59 LIMITING REACTANTS S Balance the reaction of Aluminum with Oxygen: 4 Al (s) + 3 O2 (g) → 2 Al2O3 How many grams of Al (of course in excess O2) are needed to produce 3.7 moles of Al2O3 ? 3.7 mole AL2O3 x 4 mol Al x 26.98 g Al 2 AL2O3 1 mol Al 199 g Al 2.0 X 102 g Al LIMITING REACTANTS S 4Al (s) + 3O2 (g) → 2Al2O3 If you have only 5.0 moles of O2 gas, can you still produce 3.7 moles of Al2O3? 5.0 mole O2 x 2 mol AL2O3 = 3.3 mol AL2O3 3 mol O2 No!! 60 Iron combines with Sulfur according to the following reaction: Fe (s) + S S(l) → FeS (s) In the experiment, 7.62 g Fe are allowed to react with 8.67 g S. A. Determine which of the reactants is the limiting reactant. B. Calculate the mass of FeS formed. C.How C. How much excess would remain? AMOUNT OF EXCESS S Find how much excess (in grams) would we need to react with the full amount of the limited. We subtract the excess used (in grams) from excess started with (in grams) . 61 S Grams 2 (/) Molar Mass 2 Grams 1 (/) Molar Mass 1 moles 1 moles N2 mole ratio moles moles 12 mole ratio moles moles 12 moles Product 22 moles Product 22 How many grams of Hydrogen are required to react with 100.00g of Oxygen in the production of water? If you started with 15g of Hydrogen and it is left over after the reaction stops, how much would be left? S 2H2 + O2 → 2H2O 100.0 g O2 1 mol O2 2 mol H2 2.02 g H2 = 12.60 g H2 32.0 g O2 1 mol O2 1 mol H2 15.0 g H2 – 12.6 g H2 = 2.4g H2 EXCESS 62 LIMITING REACTANTS S In the experiment, 7.62 g Fe are allowed to react with 8.67 g S. Fe (s) + S(l) → FeS (s) Start with Fe: 7.62 g Fe 1 mol Fe 55.85 g Fe 1 mol FeS 1 mol Fe = .136 mol FeS Now S: 8.67g S 1 mol S 1 mol FeS 32.05 g S 7.62 g Fe 1 mol Fe 55.85 g Fe 1 mol S 1 mol S 1 mol Fe = 0.27 mol FeS 32.05 g S = 4.4 g S 1 mol S 8.67 g S – 4.4 g S = 4.3 g S EXCESS LIMITING REACTANTS S 10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. QUESTION #1: Which reactant is limiting, which is in excess, and how much product is produced? QUESTION #2: How much of the excess is left over. 63 LIMITING REACTANTS S 10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced? 2 Al + 3 Cl2 → 2 AlCl3 Start with Al: 10.0 g Al 1 mol Al 2 mol AlCl3 = .37 mol AlCl3 Now Cl2: 27.0 g Al 2 mol 35.0g Cl2 1 mol Cl2 2 mol AlCl3 71.0 g Cl2 35g Cl2 71g Cl2 2 mol Al 3 mol Cl2 3 mol Cl2 = .32 mol AlCl3 27.0 g Al 1 mol Al = 8.8 g Al USED! 10.0 g Al – 8.8 g Al = 1.2 g Al EXCESS LIMITING REACTANTS S 15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant is limiting and how much product is made. How much excess reactant is left over? 2 K + I2 → 2 KI 64 LIMITING REACTANTS 15.0 g of potassium reacts with 15.0 g of 2 K + I2 → 2 KI iodine. 15.0 g K 1 mol K 2 mol KI = 0.384mol KI 39.1 g K 2 mol K 15.0 g I2 1 mol I2 254 g I2 15.0 g I2 1 mol I2 S 2 mol KI = 0.12 mol KI 1 mol I2 2 mol K 254.00 g I2 1 mol I2 39.10 g K 1 mol K = 4.62 g K USED! 15.0 g K – 4.62 g K = 10.38 g K EXCESS S PERCENT YIELD actual amount of a product expressed as a % of the theoretical amount of that substance 65 PERCENT YIELD % Yield = S Actual amount of product from the lab Expected calculated amount from stoichiometry PERCENT YIELD S Actual yield is the amount of product actually recovered from an experiment Theoretical (possible) yield is the maximum amount of product that could be produced from the reactant. Percent Yield is the actual yield compared to the maximum (theoretical yield) possible. 66 PERCENT YIELD S You prepared cookie dough to make 5 dozen cookies. The phone rings while a sheet of 12 cookies is baking. You talk too long and the cookies burn. You throw them out (or give them to your dog.) The rest of the cookies are okay. How many cookies could you have made (Theoretical yield)? yield)? How many cookies did you actually make to eat? (Actual yield) PERCENT YIELD S What is the percent yield of cookies? Percent Yield = Actual Yield (g)_ X 100 Theoretical Yield (g) % cookie yield = 48 cookies x 100 = 80% 60 cookies 67 PERCENT YIELD S FeS was calculated to be 12.0 g. In a lab experiment the actual yield was 10.55 g. Calculate the % Yield. 10.55g x 100 = 87.92 % 12.0 g PERCENT YIELD S 454g of NH4NO3 → NO4 + 2H2O If 454 g of NH4NO3 reacts completely How much N2O is formed if 204 g of water is produced as well? Total mass of reactants = total mass of products 454 g NH4NO3 = ___ g N2O + 204 g H2O mass of N2O = 250. g C.O.M. 68 PERCENT YIELD S 454g of NH4NO3 → NO4 + 2H2O a. Calculate the % YIELD b. If you isolated only 131 g of N2O, what is the percent yield? c. This compares the theoretical (250. g) and actual (131 g) yields. 454g of NH4NO3 → NO4 + 2H2O S Calculate the percent yield % yield = actual yield • 100% theoretical yield % yield = 131 g • 100% = 52.4% 250. g % YIELD 69 PERCENT YIELD S If we repeated the previous experiment and improved our yield from 52.4% to 90% how many grams would that be? 250. g N2O x .90 = 225 g N2O PERCENT YIELD S Mix 5.40 g of Al with 8.10 g of Cl2. What mass of Al2Cl6 can form? 2 Al + 3 Cl2 → Al2Cl6 Grams Cl2 grams Al2Cl6 Moles Cl2 1 mol Al2Cl6 3 mol Cl2 moles Al2Cl6 70 PERCENT YIELD S IF YOU MIX 5.40g OF Al WITH 8.10g OF Cl2. WHAT IS THE THEORETICAL YIELD OF Al2Cl6. 2 Al + 3 Cl2 → 5 . 40 g Al • Al2Cl6 1 mol = 0.200 mol Al 27.0 g .200 mol Al x (1 mole Al2Cl6 / 2 mol Al) = .100 mol Al2Cl6 8.10 g Cl 2 • 1 mol = 0.114 mol Cl 2 70.9 g 0.114 mol Cl2 x (1 mole Al2Cl6/ 3 mol Al) = 0.038 mol Al2Cl6 Limiting Reactant = Cl2 PERCENT YIELD 2 Al + 3 Cl2 S → 10.1 g Al2Cl6 How many grams of Al2Cl6 will form? 10.1 g Al2Cl6 8.10 g Cl2 1 Mol Cl2 70.9 g Cl2 Horseshoe 266g Mol Al2Cl6 Mol Al2Cl6 .114 mol Cl2 Mol Al2Cl6 0.038 mol Al2Cl6 3 Mol Cl2 71 HOW MUCH OF THE EXCESS WILL REMAIN AFTER THE REACTION IS COMPLETE S Cl2 was the limiting reactant. Therefore, Al was present in excess. But how much? First find how much Al was used Based on Cl2 limiting production of Al2Cl6 Then find how much Al is in excess. Based on subtracting used from had of Al Mix 5.40 g of Al with 8.10 g of Cl2. What mass of Al remains? 2 Al + 3 Cl2 grams Al2Cl6 Mol Al2Cl6 → S Al2Cl6 grams Al used 2 Mol Al Mol Al2Cl6 Mol Al 72 2 Al + 3 Cl2 → 10.1 g Al2Cl6 S How many grams of Al is Excess? 10.1 g Al2Cl6 2.05 g Al 27.0 g Al 1 Mol Al Mol Al2Cl6 266g Mol Al2Cl6 0.038 mol Al2Cl6 Excess Al 2 Mol Al Mol Al2Cl6 0.076 mol Al = Al available - Al required = 5.4 g Al - 2 .05 g Al = 3.4 g Al in excess How many grams of H2O would be generated S if 1.00 g of Glucose reacted in excess oxygen? C6H12O6 + 6 O2 → 6 CO2 + 6 H2O Starting with 1.00 g of C6H12O6… we calculate the moles of C6H12O6… use the coefficients to find the moles of H2O… and then turn the moles of water to grams 73 S Iron combines with Sulfur according to the following reaction: Fe (s) + S(l) → S FeS (s) In the experiment, 7.62 g Fe are allowed to react with 8.67 g S. a. Determine which one of the reactants is the limiting reactant. b. Calculate the mass of FeS formed. 74 Fe (s) + S(l) → FeS (s) S 7.62g Fe x 1 mole Fe = 55.85 g Fe 0.136 mol Fe 8.67 g S x 1 mol S = 32.06 g S 0.270 mol S We don’t have to go all the way around the horseshoe to find the Limiting Reactant We can stop at moles to find which runs out first - to make it easier (or whole thing if it helps) Theoretical Yield 0.136 mol Fe x 1 mol FeS 1 mol Fe 0.270 mol S x 1 mol FeS = = S 0.136 mol FeS 0.270 mol FeS 1 mol S Or you can go all the way to grams if it keeps it straight for you 75 S Determine the percent yield if 100g is collected after reaction. 0.136 mol Fe x 1 mol FeS = 119 g FeS 1 mol FeS 110 g FeS 119 g FeS = 92.4% yield 92% yield Calculate the mass of FeS if the % yield is 82%. 119 g FeS x .82 = 15.58 g FeS 15.6 g FeS S How many grams of H2O would be generated if 1.00 grams of Glucose (C (C6H12O6) reacted in excess O? C6H12O6 + 6 O2 → 6 CO2 + 6 H2O .0056 .0333 76