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Modern Chemistry Chapter 9 Stoichiometry

Modern Chemistry Chapter 9 Stoichiometry. OBJECTIVES. Define Stoichiometry Describe the importance of the mole ratio in stoichiometric calculations Write a mole ratio relating 2 substances in a chemical equation. Mg(s) + 2HCl( aq )  MgCl 2 ( aq ) + H 2 (g).

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Modern Chemistry Chapter 9 Stoichiometry

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  1. Modern Chemistry Chapter 9Stoichiometry

  2. OBJECTIVES Define Stoichiometry Describe the importance of the mole ratio in stoichiometric calculations Write a mole ratio relating 2 substances in a chemical equation.

  3. Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g) • If you know the amount of 1 substance in a reaction, you can determine the amounts of all the other substances. • Q. How many moles of H2 are obtained from 2 molHCl? • 1 mol H2 • Q. How many moles of moles of MgCl2 would result from 4 molHCl? • 2 mol MgCl2 • How could you convert moles of these substances to masses?

  4. Modern Chemistry Chapter 9Stoichiometry • composition stoichiometry deals with the mass relationships of elements in compounds. (Ch 3) • reaction stoichiometry involves the mass relationships between reactants and products in a chemical reaction • All reaction stoichiometry calculations start with a balanced chemical equation.

  5. Types of Stoichiometry Problems Classified according to the information given in the problem and the information you are expected to find, the unknown: 1. mole to moleBoth the given and the unknown quantities are amounts in moles. 2. mole to massThe given amount is in moles and the unknown amount is in grams. 3. mass to moleThe given amount is in grams and the unknown amount is in moles. 4. mass to mass Both the given and the unknown amount is in grams.

  6. Mole Ratio • mole ratio- A conversion factor that relates the amounts in moles of any two substances involved in a chemical reaction. • Found by using the coefficients in the balanced formula equation of the reaction. 2 Al2O3 (l)  4 Al (s) + 3 O2(g) Give molar ratios: Determine the amount of moles of aluminum that can be produced from 13.0 mol of aluminum oxide:

  7. MOLAR MASS • molar mass- Equal to the mass in grams of one mole of an element or a compound. • Found by adding the individual element atomic masses from the formula of the compound. 2 Al2O3 (l)  4 Al (s) + 3 O2(g) Determine the molar masses of the reactant and products: Find the number of grams of aluminum equivalent to 26.0 mol of aluminum:

  8. Careers in Chemistry:Chemical Technician • https://www.youtube.com/watch?v=7Q7Aj-c29Z8

  9. HOMEWORK Ch 9 Sec 1 Worksheet

  10. Section Review page 301 1- The branch of chemistry that deals with mass relationships in compounds and in chemical reactions. 2 HgO 2 Hg + O2 2- a) 2 mol HgO & 2 mol HgO 2 mol Hg 1 mol O2 2 mol Hg & 2 mol Hg 2 mol HgO 1 mol O2 1 mol O2 & 1 mol O2 2 mol HgO 2 mol Hg

  11. Section Review page 301 3- It is used to convert moles of one substance into moles of another substance. 4- The formula equation MUST be BALANCED so mole ratios can be determined.

  12. Section 2 Ideal Stoichiometric Calculations • Balance the chemical equations first! • Theoretical stoichiometric calculations allow us to use chemical equations to help us plan the amounts of reactants to use without having to run the reactions in the lab tell us maximum amounts of reactants and products in an ideal conditions • HOWEVER, many reactions do not proceed such that all of the reactants are converted into products. • Time to PRACTICE stoichiometric problems!!!

  13. Using Conversion Factors Amount coefficient amount of of given x of unknown = unknown substance coefficient substance in moles of known in moles # moles x mole ratio = #moles of given unknown Do practice problems #1 & #2 on page 306 of text.

  14. Page 306 #1 3 H2 + N2 2 NH3 6 mol H2 x 2 mol NH3 = 4mol NH3 3 mol H2

  15. Page 306 #2 2 KClO3 2 KCl + 3 O2 15 mol O2 x 2 mol KClO3= 10 mol KClO3 3 mol O2

  16. Chapter 9 quiz #1- mole to mole calculations 6 NaBr + Mg3(PO4)2 2 Na3PO4 + 3 MgBr2 Use the above balanced formula equation to solve the following: 1- 7.0 moles of NaBr will produce ? moles Na3PO4 2- 3.0 moles of Mg3(PO4)2 will yield ? moles MgBr2 3- 0.5 moles of NaBr will react with ? moles Mg3(PO4)2 4- 2.5 moles of NaBr will yield ? moles Na3PO4 5- 2.5 moles NaBr will produce ? moles MgBr2

  17. 6 NaBr + Mg3(PO4)2 2 Na3PO4 + 3 MgBr2 1- 7 mol NaBr x 2 mol Na3PO4 = 2.3 mol Na3PO4 6 mol NaBr 2- 3 mol Mg3(PO4)2 x 3.0 mol MgBr2 = 9 mol MgBr2 1 mol Mg3(PO4)2 3- 0.5 mol NaBr x 1 mol Mg3(PO4)2 = 0.083 mol 6 mol NaBr 4- 2.5 mol NaBr x 2 mol Na3PO4 = 0.83 mol Na3PO4 6 mol NaBr 5- 2.5 mol NaBr x 3 mol MgBr2 = 1.25 mol MgBr2 6 mol NaBr

  18. Using Conversion Factors amount of mass in given x moles unknown x molar = grams of substance moles known mass of unknown in moles unknown substance # moles x mole x molar = mass of unknown given ratio mass (in grams) unknown Do practice problems #1 & #2 on page 308 of the textbook.

  19. Page 308 #1 & 2 2 Mg + O2 2 MgO 2.00 mol Mg x 2 mol MgO x 40.3 g MgO = 80.6 g MgO 2 mol Mg 1 mol MgO 6 CO2 + 6 H2O  C6H12O6 + 6 O2 10 mol CO2 x 1 mol C6H12O6 x 180 g C6H12O6 = 300 g C6H12O6 6 mol CO2 1 mol C6H12O6

  20. Chapter 9 Quiz #2- mole-mass problems 3 MgF2 + Al2O3 3 MgO + 2 AlF3 Use the above balanced formula equation to answer the following questions. 1- 2.0 mol MgF2 will yield ? grams of MgO 2- 4.0 mol of Al2O3  ? grams of AlF3 3- If 6.0 mol of MgO is produced, ? grams of AlF3 4- 0.6 mol MgF2  ? grams of AlF3 5- 2.75 mol Al2O3  ? grams of MgO

  21. 3 MgF2 + Al2O3 3 MgO + 2 AlF3 1- 2.0 mol MgF2 x 3 mol MgO x 40.3 g MgO = 80.6 g 3 mol MgF2 1 mol MgO MgO 2- 4.0 mol Al2O3 x 2 mol AlF3x 84.0 g AlF3= 672.0g 1 mol Al2O3 1 mol AlF3 AlF3 3- 6.0 mol MgO x 2 mol AlF3 x 84.0 g AlF3 = 336 g AlF3 3 mol MgO 1 mol AlF3 4- 0.6 mol MgF2 x 2 mol AlF3 x 84.0 g AlF3 = 33.6 g AlF3 3 mol MgF2 1 mol AlF3 5- 2.75 mol Al2O3 x 3 mol MgO x 40.3 g MgO = 332gMgO 1 mol Al2O3 1 mol MgO

  22. Using Conversion Factors mass (g)x 1 mol given x mol unknown = moles of of given molar mass mol given unknown substance of given substance grams x 1 x mole ratio = moles unknown molar mass Do practice problems #1 & #2 on page 309 of the textbook.

  23. Practice problems page 309 2 HgO 2 Hg + O2 125 g O2 x 1 mol O2 x 2 mol HgO = 7.81 mol HgO 32 g O2 1 mol O2 125 g O2 x 1 mol O2 x 2 mol Hg = 7.81 mol Hg 32 g O2 1 mol O2

  24. chapter 9 quiz #3- mass-mole problems Na2O + CaF2 2 NaF + CaO Use the above equation to solve the problems. 1- 156.1 grams of CaF2  ? mol CaO 2- 186 g Na2O  ? mol NaF 3- 31 g Na2O  ? mol CaO 4- 31 g Na2O  ? mol NaF 5- A yield of 84 g NaF  ? mol CaO

  25. Na2O + CaF2 2 NaF + CaO 1- 156.1 g CaF2 x 1 mol CaF2 x 1 mol CaO = 2.00 mol CaO 78.1 g CaF2 1 mol CaF2 2- 186.0 g Na2O x 1 mol Na2O x 2 mol NaF = 6.0 mol NaF 62 g Na2O 1 mol Na2O 3- 31.0 g Na2O x 1 mol Na2O x 1 mol CaO = 0.5 mol CaO 62.0 g Na2O 1 mol Na2O 4- 31.0 g Na2O x 1 mol Na2O x 2 mol NaF = 1.0 mol NaF 62.0 g Na2O 1 mol Na2O 5- 84.0 g NaF x 1 mol NaF x 1 mol CaO = 1.0 mol CaO 42.0 g NaF 2 mol NaF

  26. Chemistry Chapter 9- Stoichiometry Practice Problems 2 NaF + CaO  Na2O + CaF2 1- 4.5 moles of NaF will produce -?- moles of Na2O ? 4.5 mol NaF x 1 mol Na2O/2 mol NaF = 2.25 moles Na2O 2- 3.2 moles of CaO will produce -?- grams of CaF2 ? 3.2 mol CaO x 1 mol CaF2/1 mol CaO x 78.1 g CaF2/mol CaF2 = 249.9 g CaF2

  27. 2 NaF + CaO  Na2O + CaF2 3- 168.0 grams of NaF will produce -?- moles of Na2O ? 168.0 /42.0 x 1 /2 = 2.0 mol Na2O 4- 112.2 grams of CaO will produce -?- moles of CaF2 ? 112.2/56.1 x 1/1 = 2.0 mol CaF2

  28. 5- Calculate the molar mass of each of the reactants & products of the above balanced formula equation. Use the molar masses in the following problems. a- AlN = (1 x 27.0) + (1 x 14.0) = 41.0 g/mol b- Na2O = (2 x 23.0) + (1 x 16.0) = 62.0 g/mol c- Al2O3 = (2 x 27.0) + (3 x 16.0) = 102.0 g/mol d- Na3N = (3 x 23.0) + (1 x 14.0) = 83.0 g/mol

  29. 2 AlN + 3 Na2O  Al2O3 + 2 Na3N 6- 82.0 grams of AlN will produce -?- moles of Al2O3 ? 82.0/41.0 x 1/2 = 1.0 mol Al2O3 7- 164.0 grams AlN will produce -?- moles of Na3N ? 164.0/41.0 x 2/2 = 4.0 mol Na3N

  30. 2 AlN + 3 Na2O  Al2O3 + 2 Na3N 8- 2.5 moles of Na2O will produce -?- grams of Na3N ? 2.5 x 2/3 x 83.0 = 138.3 g Na3N 9- 0.75 moles of Na2O will produce -?- moles of Al2O3 ? 0.75 x 1/3 = 0.25 mol Al2O3 10- 11.0 moles of Na2O will produce -?- grams of Na3N ? 11.0 x 2/3 x 83.0 = 608.7 g Na3N

  31. 2 H2 + O2 2 H2O 11- 8.0 grams of H2 will react with -?- moles of O2 ? 8.0/2.0 x 1/2 = 2.0 mol O2 12- 64.0 grams of O2 will produce -?- moles of H2O ? 64.0/32.0 x 2/1 = 4.0 mol H2O 13- 0.25 moles of H2 will produce -?- moles of H2O ? 0.25 x 2/2 = 0.25 mol H2O

  32. 2 H2 + O2 2 H2O 14- 1.5 moles of H2 will produce -?- grams of H2O ? 1.5 x 2/2 x 18.0 = 27.0 g H2O 15- 14 moles of O2 will produce -?- moles of H2O ? 14 x 2/1 = 28.0 mol H2O

  33. Using Conversion Factors mass (g) x 1 molgiven x mol unknown x molar mass unknown = mass of given molar mass molgiven 1 molunknown of un- substance of given known grams x 1 x mole ratio x molar mass = mass of unknown given molar mass Do practice problems #1, #2, & #3 on page 311 of textbook.

  34. Practice problems page 311 NH4NO3 N2O + 2 H2O 32 g N2O x 1 mol N2O x 1 mol NH4NO3 x 80 g NH4NO3 = 60 g NH4NO3 44 g N2O 1 mol N2O 1 mol NH4NO3

  35. HOMEWORKSec 2 WS

  36. Chapter 9 quiz #4- mass-mass problems Na2O + CaF2 2 NaF + CaO 1- 124 g Na2O  ? grams NaF 2- 124 g Na2O  ? grams CaO 3- 234.3 g CaF2  ? g NaF 4- 234.3 g CaF2  ? g CaO 5- 84.0 g NaF  ? g CaO

  37. Na2O + CaF2 2 NaF + CaO 1- 124 g Na2O x 1 mol Na2O x 2 mol NaF x 42.0 g NaF = 168 g 62 g Na2O 1 mol Na2O 1 mol NaF 2- 124 g Na2O x 1 mol Na2O x 1 mol CaO x 56.1 g CaO = 112.2 g 62 g Na2O 1 mol Na2O 1 mol CaO 3- 234.3 g CaF2 x 1 mol CaF2 x 2 mol NaF x 42.0 g NaF = 252 g 78.1 gCaF2 1 mol CaF2 1 mol NaF 4- 234.3 g CaF2 x 1 mol CaF2 x 1 mol CaO x 56.1 g CaO = 168.3 g 78.1 gCaF2 1 mol CaF2 1 mol CaO 5- 84.0 g NaF x 1 mol NaF x 1 mol CaO x 56.1 g CaO = 56.1 g 42.0 g NaF 2 mol NaF 1 mol CaO

  38. The “MOLE HILL” x mole ratio # moles known # moles unknown ÷ molar mass x molar mass of known of unknown mass of known mass of unknown

  39. Stoichiometry Practice Problems 2 H2 + O2 2 H2O 1) 2.5 mol H2 x 1 mol O2 = 1.25 mol O2 2 mol H2 2) 2.5 mol H2 x 2 mol H2O = 2.5 mol H2O 2 mol H2 3) 2.5 mol H2 x 1 mol O2 x 32 g O2 = 40 g O2 2 mol H2 1 mol O2 4) 2.5 mol H2 x 2 mol H2O x 18 g H2O = 45 g H2O 2 mol H2 1 mol H2O

  40. Stoichiometry Practice Problems 2 H2 + O2 2 H2O 5) 16 g H2 x 1 mol H2 x 1 mol O2 = 4.0 mol O2 2 g H2 2 mol H2 6) 16 g H2 x 1 mol H2 x 2 mol H2O = 8.0 mol H2O 2.0 g H2 2 mol H2 7) 16 g H2 x 1 mol H2 x 1 mol O2 x 32 g O2= 128 g O2 2.0 g H2 2 mol H2 1 mol O2 8) 16 g H2 x 1 mol H2 x 2 mol H2O x 18 g H2O = 144 g H2O 2.0 g H2 2 mol H2 1 mol H2O

  41. Section 3 Stoichiometry In the lab, a reaction is carried out with the required amount of each reactant. In many cases, one or more reactants is present in excess meaning we have more than exact amount required to react…

  42. The buns, slices of cheese, leaves of lettuce, and 5 pieces of burger meat are all reacting to produce burgers. Assuming you use one of each to make a burger, the limiting reagent will be the lettuce leaves. The "reaction" will produce 3 burgers - the amount determined by the limiting reagent - and have 1 bun, 3 slices of cheese, and 2 pieces of meat in excess.

  43. Limiting Reactant & Percentage Yield • limiting reactant is the reactant that limits the amount of the other reactant that can combine and the amount of product that can be formed in a chemical reaction. • excess reactantis the substance that is NOT completely used up in a chemical reaction.

  44. Sample & Practice Problems Silicon dioxide (quartz) is usually quite unreactive but reacts readily with hydrogen fluoride: SiO2(s) + 4HF(g)  SiF4(g) + H2O(l) If 6.0 mole HF is added to 4.5 mol SiO2, which is the limiting reactant? Known: Unknown: • Pick 1 product. • Use the given amounts of each reactant to calculate the amount of product that can be produced from each reactant. • The LR is the reactant that produces the smaller number of moles of product. • The smallest amount of product = max amt that can form

  45. Take the reaction: NH3 + O2 NO + H2O. In an experiment, 3.25 g of NH3 are allowed to react with 3.50 g of O2. a. Which reactant is the limiting reagent?   O2 b. How many grams of NO are formed?   2.63 g NO c. How much of the excess reactant remains after the reaction?   1.76 g NH3 left

  46. If 4.95 g of ethylene (C2H4) are combusted with 3.25 g of oxygen. a. What is the limiting reagent?   O2 b. How many grams of CO2 are formed?   2.98 g CO2

  47. Extra Credit: QuickLAB • Do the QuickLAB titled “Limiting Reactants in a Recipe” on page 316 of the textbook. Yes, cooking IS chemistry!

  48. Percentage Yield • theoretical yield is the maximum amount of product that can be produced from a given amount of reactant • actual yield of a product is the measured amount of a product obtained from a reaction • percentage yieldis the ratio of the actual yield to the theoretical yield multiplied by 100 • percentage yield = actual yield x 100 theoretical yield

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