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IB Topic 1: Quantitative Chemistry 1.4: Mass Relationships in Chemical Reactions

2. Review: Identify the mole ratio of any two species in a chemical reaction. . How do you calculate the quantity of reactants and products in chemical reactions?Use the chemical equation to predict ratio of reactants and productsN2 3H2?2NH31 molecule N2 reacts with 3 molecules H2 producin

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IB Topic 1: Quantitative Chemistry 1.4: Mass Relationships in Chemical Reactions

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    1. 1 IB Topic 1: Quantitative Chemistry 1.4: Mass Relationships in Chemical Reactions Calculate theoretical yields from chemical equations. Determine the limiting reactant and the reactant in excess when quantities of reacting substances are given. Solve problems involving theoretical, experimental and percentage yield. Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas. Solve problems using the ideal gas equation, PV = nRT Apply Avogadros law to calculate reacting volumes of gases Apply the concept of molar volume at standard temperature and pressure in calculations Analyze graphs relating to the ideal gas equation

    2. 2 Review: Identify the mole ratio of any two species in a chemical reaction. How do you calculate the quantity of reactants and products in chemical reactions? Use the chemical equation to predict ratio of reactants and products N2 + 3H2 ? 2NH3 1 molecule N2 reacts with 3 molecules H2 producing 2 molecules NH3 1 mole N2 reacts with 3 moles H2 producing 2 moles NH3 The balanced equations gives us the ratio in particles or moles (not mass) of the chemicals involved in the reaction. The ratio of N2 to NH3 is 1:2 The ratio of NH3 to H2 is 2:3 The ratio of H2 to N2 is 3:1

    3. 3 Review: Identify the mole ratio of any two species in a chemical reaction. Consider the reaction: 2C8H18 + 25O2 ? 16CO2 + 18H2O What is the ratio of O2 to CO2? 25:16 What is the ratio of H2O to C8H18? 18:2 What is the ratio of CO2 to H2O? What is the ratio of O2 to C8H18? Remember these ratios are in particles or moles. We will be using moles.

    4. 4 Calculate theoretical yields from chemical equations. Theoretical Yield is the amount of product that would result if all the limiting reagent reacted. In essence, theoretical yield is what you calculate We use a process called stoichiometry: The calculation of quantities in chemical reactions

    5. 5 Mole-Mole Conversions 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

    6. 6 Mole-Mass Conversions 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

    7. 7 Practice Calculate the mass in grams of Iodine required to react completely with 0.50 moles of aluminum. 2 Al + 3 I2 ? 2 AlI3

    8. 8 Mass-Mole 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

    9. 9 Practice Calculate how many moles of oxygen are required to make 10.0 g of aluminum oxide 4 Al + 3 O2 ? 2 Al2O3

    10. 10 Mass-Mass Conversions Most often we are given a starting mass and want to find out the mass of a product we will get (called theoretical yield) or how much of another reactant we need to completely react with it (no leftover ingredients!) Now we must go from grams to moles, mole ratio, and back to grams of compound we are interested in

    11. 11 Mass-Mass Conversion Ex. Calculate how many grams of ammonia are produced when you react 2.00g of nitrogen with excess hydrogen. N2 + 3 H2 ? 2 NH3

    12. 12 1.4.1 Calculate theoretical yields from chemical equations. How much oxygen does it take to burn 100.0 grams of octane?

    13. 13 1.4.1 Calculate theoretical yields from chemical equations. How much oxygen does it take to burn 100.0 grams of octane? Write the balanced chemical equation: 2C8H18 + 25O2 ? 16CO2 + 18H2O

    14. 14 1.4.1 Calculate theoretical yields from chemical equations. How much oxygen does it take to burn 100.0 grams of octane? 2C8H18 + 25O2 ? 16CO2 + 18H2O Mass of O2 required: 10.97 mol (32.00 g/mol) = 351.0 g O2

    15. 15 1.4.1 Calculate theoretical yields from chemical equations. In a spectacular reaction called the thermite reaction, iron(III) oxide reacts with aluminum producing iron and aluminum oxide. How many grams of iron will be produced from 43.7 grams of aluminum?

    16. 16 1.4.1 Calculate theoretical yields from chemical equations. In a spectacular reaction called the thermite reaction, iron(III) oxide reacts with aluminum producing iron and aluminum oxide. How many grams of iron will be produced from 43.7 grams of aluminum? Write the balanced chemical equation: Fe2O3 + 2Al ? 2Fe + Al2O3

    17. 17 1.4.1 Calculate theoretical yields from chemical equations. In a spectacular reaction called the thermite reaction, iron(III) oxide reacts with aluminum producing iron and aluminum oxide. How many grams of iron will be produced from 43.7 grams of aluminum? Write the balanced chemical equation: Fe2O3 + 2Al ? 2Fe + Al2O3 Mass of Fe produced: 1.62 mol (55.8 g/mol) = 90.4 g Fe

    18. 18 Limiting Reactant: Cookies 1 cup butter 1/2 cup white sugar 1 cup packed brown sugar 1 teaspoon vanilla extract 2 eggs 2 1/2 cups all-purpose flour 1 teaspoon baking soda 1 teaspoon salt 2 cups semisweet chocolate chips Makes 3 dozen

    19. 19 1.4.2 Determine the limiting reactant and the reactant in excess when quantities of reacting substances are given. Limiting Reactants You are given amounts for two reactants and one reactant will run out first. This is called the limiting reactant. The reactant that is left over is called the excess reactant. For example: A strip of zinc metal weighing 2.00 g is placed in a solution containing 2.50 g of silver nitrate causing the following reaction to occur: Zn + 2AgNO3 ? 2Ag + Zn(NO3)2 How many grams of Ag will be produced? AgNO3 is the limiting reactant because it will produce the least amount of Ag. Zn is the reactant in excess. There will be Zn left over after the reactionAgNO3 is the limiting reactant because it will produce the least amount of Ag. Zn is the reactant in excess. There will be Zn left over after the reaction

    20. 20 Limiting Reactant 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.

    21. 21 Limiting Reactant To find the correct answer, we have to try all of the reactants. We have to calculate how much of a product we can get from each of the reactants to determine which reactant is the limiting one. The lower amount of a product is the correct answer. The reactant that makes the least amount of product is the limiting reactant. Once you determine the limiting reactant, you should ALWAYS start with it! Be sure to pick a product! You cant compare to see which is greater and which is lower unless the product is the same!

    22. 22 Limiting Reactant: Example 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: Now Cl2:

    23. 23 LR Example Continued We get 49.4g of aluminum chloride from the given amount of aluminum, but only 43.9g of aluminum chloride from the given amount of chlorine. Therefore, chlorine is the limiting reactant. Once the 35.0g of chlorine is used up, the reaction comes to a complete .

    24. 24 Limiting Reactant Practice 15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant is limiting and how much product is made.

    25. 25 Limiting Reactant Practice 2 K + I2 ? 2 KI Potassium: Iodine: Iodine is the limiting reactant and we get 19.6 g of potassium iodide

    26. 26 Finding the Amount of Excess By calculating the amount of the excess reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess. Can we find the amount of excess potassium in the previous problem?

    27. 27 Finding Excess Practice 15.0 g of potassium reacts with 15.0 g of iodine. 2 K + I2 ? 2 KI We found that Iodine is the limiting reactant, and 19.6 g of potassium iodide are produced.

    28. 28 Limiting Reactant: Recap You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT. Convert ALL of the reactants to the SAME product (pick any product you choose.) The lowest answer is the correct answer. The reactant that gave you the lowest answer is the LIMITING REACTANT. The other reactant(s) are in EXCESS. To find the amount of excess, subtract the amount used from the given amount. If you have to find more than one product, be sure to start with the limiting reactant. You dont have to determine which is the LR over and over again!

    29. 29 Practice How many grams of calcium nitride are produced when 2.00 g of calcium reacts with an excess of nitrogen? 3 Ca + N2 ? Ca3N2

    30. 30 1.4.3 Solve problems involving theoretical, experimental and percentage yield.

    31. 31 1.4.3 Solve problems involving theoretical, experimental and percentage yield. Theoretical Yield is the amount of product that would result if all the limiting reagent reacted (i.e. the calculated amount) Actual Yield is the amount of product actually obtained from a reaction in lab

    32. 32 1.4.3 Solve problems involving theoretical, experimental and percentage yield. The theoretical yield from a chemical reaction is the yield calculated by assuming the reaction goes to completion. In practice we often do not obtain as much product from a reaction mixture as theoretically possible due to a number of factors: Many reactions do not go to completion Some reactants may undergo two or more reactions simultaneously forming undesired products Not all of the desired product can be separated from the rest of the products The amount of a specified pure product actually obtained from a given reaction is the experimental or actual yield.

    33. 33 1.4.3 Solve problems involving theoretical, experimental and percentage yield. Percentage yield indicates how much of a desired product is obtained from a reaction

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    35. 35 1.4.3 Solve problems involving theoretical, experimental and percentage yield. Consider the reaction: P4O10 + 6H2O ? 4H3PO4 If 558 g of P4O10 reacts, the experimental yield of H3PO4 is 746 g. What is the percentage yield of H3PO4? Theoretical amt: 558 g/284 g/mol = 1.965 mol P4O10 X H3PO4 = 4 H3PO4 = 7.86 mol H3PO4 1.965 P4O10 1 P4O10 7.86 mol H3PO4 x 98.0 = 770. g H3PO4 Percentage yield: 746 g x 100% = 96.9% 770 g

    36. 36 1.4.3 Solve problems involving theoretical, experimental and percentage yield. Consider the reaction: P4 + 5O2 ? P4O10 If 272 g of phosphorus reacts, the percentage yield of tetraphosphorus decoxide is 89.5%. What mass of P4O10 is obtained? Theoretical amt: 272 g/124.0 g/mol = 2.194 mol P4 X P4O10 = 1 P4O10 = 2.194 mol P4O10 2.19 P4 1 P4 2.194 mol P4O10 x 284.0 = 623 g Experimental amt: 623 g x .895 = 558 g

    37. 37 1.4.3 Solve problems involving theoretical, experimental and percentage yield. Consider the reaction: Fe2O3 + 3CO ? 2Fe + 3CO2 When 84.8 g iron(III) oxide reacts with an excess of carbon monoxide, 54.3 g of iron is produced. What is the percentage yield of this reaction?

    38. 38 1.4.3 Solve problems involving theoretical, experimental and percentage yield. Consider the reaction: Fe2O3 + 3CO ? 2Fe + 3CO2 When 84.8 g iron(III) oxide reacts with an excess of carbon monoxide, 54.3 g of iron is produced. What is the percentage yield of this reaction? Theoretical amt: 84.8 g/159.2 g/mol = 0.5327 mol Fe2O3 X Fe = 2 Fe = 1.065 mol Fe 0.5327 mol Fe2O3 1 Fe2O3 1.065 mol Fe x 55.8 g.mol = 59.4 g Fe Percentage yield: 54.3 g x 100% = 91.4% 59.4 g

    39. 39 1.4.3 Solve problems involving theoretical, experimental and percentage yield. If 50.0 g silicon dioxide is heated with an excess of carbon, the percentage yield of silicon carbide 76.0%. What mass of silicon carbide will actually be produced? SiO2 + 3C ? SiC + 2CO

    40. 40 1.4.3 Solve problems involving theoretical, experimental and percentage yield. If 50.0 g silicon dioxide is heated with an excess of carbon, the percentage yield of silicon carbide 76.0%. What mass of silicon carbide will actually be produced? SiO2 + 3C ? SiC + 2CO Theoretical amt: 50.0 g/60.1 g/mol = 0.8319 mol SiO2 X SiC = 1 SiC = 0.8319 mol SiC 0.8319 mol SiO2 1 SiO2 0.8319 mol SiC x 40.1 g.mol = 33.4 g SiC Experimental yield: 33.4 g x .760% = 25.4 g SiC Read pg. 256-257. pg. 259: 30,32

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    42. 42 1.4.6 Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas. Variables That Describe a Gas Pressure (P) measured in kPa, mm Hg, atm 101.3 kPa = 760 mm Hg = 1.00 atm Volume (V) measured in dm3 or L 1 dm3 = 1000 cm3 = 1 L = 1000 mL Temperature (T) measured in K (Kelvin) K = oC + 273 Amount of matter (n) measured in moles

    43. 43 Gaseous Volume Relationships in Chemical Reactions Kinetic Theory: Tiny particles in all forms of matter are in constant motion Application to Gases A gas is composed of particles that are considered to be small, hard spheres that have insignificant volume and are relatively far apart from one another. Between the particles there is empty space. No attractive or repulsive forces exist between the particles. 2) The particles in a gas move rapidly in constant random motion. They travel in straight paths and move independently of each other. They change direction only after a collision with one another or other objects. 3) All collisions are perfectly elastic. Total kinetic energy remains constant.

    44. 44 1.4.6 Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas. Gas Laws: Boyles Law Relates Pressure-Volume As pressure increases, volume decreases if temperature and amount remain constant. Spaces between particles so particles can move close closer together P1 x V1 = P2 x V2 See pg 335: Sample problem 12-1. Do practice problems pg 335: 10,11

    45. 45 1.4.6 Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas. Gas Laws: Charless Law Relates Temperature-Volume As temperature increases, volume increases if pressure and amount remain constant Particles gain kinetic energy, move farther apart V1/T1 = V2/T2 ; T has to be in Kelvin ; K= oC + 273 See pg 337: Sample problem 12-2. Do practice problems pg 337: 12,13

    46. 46 1.4.6 Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas. Gas Laws: Gay-Lussacs Law Relates Temperature-Pressure As temperature increases, pressure increases if volume and amount remain constant. Particles gain kinetic energy so they move faster and have more collisions P1/T1 = P2/T2 ; T has to be in Kelvin ; K = oC + 273 See pg 338: Sample problem 12-3. Do practice problems pg 338-339: 14,15

    47. 47 1.4.6 Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas. Gas Laws: Combined Gas Law Relates Temperature-Pressure-Volume P1 x V1/T1 = P2 x V2/T2 T has to be in Kelvin See pg 340: Sample problem 12-4. Do practice problems pg 340: 16,17

    48. 48 1.4.7 Solve problems using the ideal gas equation, PV = nRT Gas Laws: Ideal Gas Law Relates Temperature-Pressure-Volume-Amount PV = nRT P = pressure in kPa V = volume in dm3 or L n = moles R = Gas constant (8.31) T = temperature in K Ideal gas: particles have no volume and are not attracted to each other Pg 342: Sample 12-5. Practice 22,23 Pg 343: Sample 12-6. Practice 24,25

    49. 49 1.4.4 Apply Avogadros law to calculate reacting volumes of gases Avogadros Hypothesis: Equal volumes of gases at the same temperature and pressure contain equal numbers of particles (moles). At STP (standard temperature & pressure: 273 K and 101.3 kPa) 1 mole of any gas occupies a volume of 22.4 dm3 (L). Read pg.347-348. Pg. 348-349: Practice 31-36

    50. 50 1.4.4 Apply Avogadros law to calculate reacting volumes of gases Assuming STP, how many dm3 of oxygen are needed to produce 19.8 dm3 SO3 according to: 2SO2(g) + O2(g) ? 2SO3(g) Since equal volumes of gases contain the same number of moles, we can use the equation coefficients with the volumes. X dm3 O2 = 1 O2 = 9.90 dm3 O2 19.8 dm3 SO2 2 SO2

    51. 51 1.4.4 Apply Avogadros law to calculate reacting volumes of gases Nitrogen monoxide and oxygen combine to form the brown gas nitrogen dioxide. How many cm3 of nitrogen dioxide are produced when 3.4 cm3 of oxygen reacts with an excess of nitrogen monoxide? Assume conditions of STP.

    52. 52 1.4.4 Apply Avogadros law to calculate reacting volumes of gases Nitrogen monoxide and oxygen combine to form the brown gas nitrogen dioxide. How many cm3 of nitrogen dioxide are produced when 3.4 cm3 of oxygen reacts with and excess of nitrogen monoxide? Assume conditions of STP. 2NO + O2 ? 2NO2 X cm3 NO2 = 2 NO2 = 6.8 cm3 O2 3.4 cm3 O2 1 O2 Pg. 249: 15-16. Pg. 250: 17,18

    53. 53 1.4.5 Apply the concept of molar volume at standard temperature and pressure in calculations Tin(II) fluoride, formerly found in many kinds of toothpaste, is formed by this reaction: Sn(s) + 2HF(g) ? SnF2(s) + H2(g) How many dm3 of HF are needed to produce 9.40 dm3 H2 at STP? How many grams of Sn are needed to react with 20.0 dm3 HF at STP? What volume of H2 is produced from 37.4 g Sn?

    54. 54 1.4.5 Apply the concept of molar volume at standard temperature and pressure in calculations Tin(II) fluoride, formerly found in many kinds of toothpaste, is formed by this reaction: Sn(s) + 2HF(g) ? SnF2(s) + H2(g) How many dm3 of HF are needed to produce 9.40 dm3 H2 at STP? x HF = 2 HF = 18.8 dm3 HF 9.40 dm3 H2 1 H2

    55. 55 1.4.5 Apply the concept of molar volume at standard temperature and pressure in calculations Tin(II) fluoride, formerly found in many kinds of toothpaste, is formed by this reaction: Sn(s) + 2HF(g) ? SnF2(s) + H2(g) How many grams of Sn are needed to react with 20.0 dm3 of HF at STP? 20.0 dm3 HF / 22.4 dm3mol-1 = .893 mol HF x Sn = 1 Sn = .447 mol Sn .893 HF 2 HF .893 mol Sn x 118.69 = 53.0 g

    56. 56 1.4.5 Apply the concept of molar volume at standard temperature and pressure in calculations Tin(II) fluoride, formerly found in many kinds of toothpaste, is formed by this reaction: Sn(s) + 2HF(g) ? SnF2(s) + H2(g) What volume of H2 at STP is produced from 37.4 g Sn? 37.4 g / 118.69 gmol-1 = .315 mol Sn x H2 = 1 H2 = .315 mol H2 .315 Sn 1 Sn .315 mol H2 x 22.4 dm3 = 7.06 dm3

    57. 57 1.4.8 Analyze graphs relating to the ideal gas equation

    58. 58 1.4.8 Analyze graphs relating to the ideal gas equation Real gases deviate from ideal behavior at low and high pressures and temperatures. Gas molecules do have some attraction for each other Gas molecules have a volume

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