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# Chapter 8

Chapter 8. Quantities In Reactions. Homework. Assigned Problems (odd numbers only) “Problems” 17 to 73 “Cumulative Problems” 75-95 “Highlight Problems” 97-99. Calculations Using Balanced Equations: Stoichiometry.

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## Chapter 8

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1. Chapter 8 Quantities In Reactions

2. Homework • Assigned Problems (odd numbers only) • “Problems” 17 to 73 • “Cumulative Problems” 75-95 • “Highlight Problems” 97-99

3. Calculations Using Balanced Equations: Stoichiometry • Stoichiometry is the study of the quantitative relationships among reactants and products in a chemical reaction • These chemical calculations can be used to determine the amount of one reactant needed to completely react with another • Or, to determine the amount of reactant needed to produce a desired amount of product • To calculate chemical quantities in reactions involves knowing how to interpret a balanced chemical equation

4. Calculations Using Balanced Equations: Law of Conservation of Mass • As a chemical reaction proceeds: • Reactants are consumed and new materials with new chemical properties are produced • Bonds are broken, formed, or atoms are rearranged which produces new substances • No material is lost or gained as original substances change to new substances

5. Calculations Using Balanced Equations: Law of Conservation of Mass • Law of Conservation of Mass • Quantity of matter does not change during a chemical reaction • The sum of the masses of products is equal to the sum of masses of reactants • Atoms are neither created nor destroyed in chemical reactions • Only a balanced equation obeys this law

6. Mole Relationships in Chemical Equations:Conservation of Mass • A balanced equation has the same number of atoms on each side of the arrow 2 2

7. Information Available from a Balanced Chemical Equation • “1 mole of methane gas reacts with 2 moles of oxygen gas to produce 1 mole of carbon dioxide and 2 moles of water vapor.” • Multiplying each of the molar masses by the coefficient will give the total mass of reactants and products

8. Making Molecules:Mole-Mole Conversions A balanced chemical equations tell us: • The formulas and symbols of the reactants and products • The physical state of each substance • If special conditions such as heat are required • The number of molecules, formula units, or atoms of each type of molecule involved in the reaction • Number can be in terms of single atoms, or moles of atoms • The relative number of moles of each reactant and product

9. Making Molecules:Mole-Mole Conversions • In a balanced equation, conversion from moles of one substance to another will be determined by the values of the coefficients • Balancing an equation will generate the coefficients that equal the number of moles of each reactant and product • To determine how many moles of methanol would be produced if 0.295 moles of hydrogen gas is consumed: • Requires a mole to mole relationship between methanol and hydrogen gas 2

10. Making Molecules:Mole-Mole Factors • Converting the given mole amount of hydrogen gas enables you to find the number of moles methanol produced • Coefficients of the balanced equation can be used to make mole to mole relationships between the different reactants and products 2 Find: mol CH3OH Given: 0.295 mol H2 2 mol H2 = 1 mol CH3OH

11. Making Molecules:Mole-Mole Factors The unit path begins with moles of H2 and ends with moles of CH3OH From any mole-mole relationship, two mole-mole conversion factors can be made: 2 Mole-mole factor mol H2 mol CH3OH 2 mol H2 = 1 mol CH3OH

12. Making Molecules:Mole-Mole Factors • Write all of the possible mole-mole factors for the following chemical equation 2 2 1 mol O2 = 2 mol H2O 2 mol H2 = 1 mol O2 2 mol H2 = 2 mol H2O

13. Using Mole-Mole Factors in Calculations:Calculating Moles of a Product • Calculations based on balanced equations require the use of mole to mole (conversion) factors • Equation must be balanced • Identify the known (given) and needed (find) substances • Make the conversion factor based on:

14. Using Mole-Mole Factors in Calculations:Calculating Moles of a Product grams B grams A MM of A MM of B Stoichiometry moles B moles A moles B moles A • Using mole-mole factors from a BALANCED chemical equation • You can convert moles of one compound to moles of another compound using the correct mole-mole factor Mole-mole factor

15. Using Mole-Mole Factors in Calculations:Calculating Moles of a Product • Calculate the moles of CO2 formed when 4.30 moles of C3H8 reacts with (the required) 21.5 moles of O2 • Balance the equation • Plan to convert the given amount of moles to the needed amount of moles • Use coefficients to state the relationships and mole-mole factors • Set up the problem using the mole-mole factor and canceling units 4 5 3

16. Using Mole-Mole Factors in Calculations:Calculating Moles of a Product 5 4 3 MM of B MM of A Mole-mole factor Given: Find: Mole-mole factor Mole-mole relation 1 mol C3H8 = 3 mol CO2

17. Using Mole-Mole Factors in Calculations:Calculating Moles of a Product • Set up the problem using the mole-mole factor that cancels given moles and provides needed moles 5 4 3 Mole-mole factor

18. Making Molecules: Mass-to-Mass Conversions • From the balanced equation • It is also possible to start with a known mass of one substance • Then convert to moles of another substance • Start with a given amount (of grams) of a substance • Then use mole-mole factor to find the sought-after mass of another substance

19. Making Molecules: Mass-to-Mass Conversions • The most common type of stoichiometric calculation is the mass-to-mass calculation • In this type of problem: • The mass of one substance involved in a chemical reaction is given • Find the mass of another substance involved in the reaction • If a chemist only has so many grams of a certain chemical • Can calculate how many grams of another substance can be produced • Can calculate how many grams of another reactant are required to react with it

20. Mass-to-Mass Conversions • To convert the grams of one substance to grams of another substance: • Find the mole-mole factor using the coefficients in the balanced equation • You can only relate (moles-moles) of two compounds, not grams-to-grams • Ratios ONLY apply to moles, NOT grams • Must convert grams to moles, then use mole-mole factor MM of B MM of A moles B Mole-mole factor moles A

21. Mass-to-Mass Conversions Mass-to-mass conversions begin with a given mass of substance A By use of the balanced equation, find the mass of another (substance B) Convert grams of A to moles of A Convert the moles of A to moles of B by use of mole-mole ratio Convert moles of B to mass of B grams B grams A MM of A MM of B Stoichiometry moles B moles A moles B moles A

22. Mass-to-Mass Conversions(Mass of Product from Mass of Reactant) grams CO2 grams C3H8 MM of C3H8 MM of CO2 Stoichiometry moles CO2 moles C3H8 moles CO2 moles C3H8 • Calculate the mass of carbon dioxide produced when 96.1 g of propane react with sufficient oxygen. • Balance the equation • Plan to convert the given mass to given moles • Convert the given moles to sought-after moles by the use of mole-mole factor • Convert the needed moles to needed mass 4 5 3 Mole-mole factor

23. Mass-to-mass Conversions: Example 1 grams CO2 grams C3H8 MM of A MM of B Stoichiometry moles CO2 moles C3H8 moles B moles A • Write the equalities • 1 mol C3H8 = 44.09 g C3H8 • 1 mol CO2 = 44.01 g CO2 • 1 mol C3H8 = 3 mol CO2 to create mole-mole factor 3 5 4 Find: g of CO2 Given: 96.1 g C3H8

24. Mass-to-Mass Conversions: Example 1 3 5 4 Xg CO2

25. Mass-to-Mass Conversions: Example 2 • What mass of carbon monoxide and what mass of hydrogen are required to form 6.0 kg of methanol by the following reaction: 2

26. Mass Calculations Example 2

27. Limiting Reactant, Theoretical Yield, and Percent Yield • The chemical reactants are usually not present in the exact mole-mole ratios as stated in the balanced chemical equation • Often, one of the reactants is purposely added in an excess amount • Reasons include: • Increase the rate of reaction • To ensure that one reactant is completely used up (reacted) • Reactants are not completely converted to products as stated on paper (theory)

28. Limiting Reactant, Theoretical Yield, and Percent Yield • Chemical reactions with two or more reactants will continue until one of the reactants is used up (consumed) • If one of the reactants is used up, the reaction will stop because there is not enough of the other reactant to react with it • The reactant used up is called the limiting reactant (reagent) • This reactant limits the amount of product that can be made

29. Limiting Reactant • When you make peanut butter sandwiches • Required: 2 slices of bread and 1 tbsp. peanut butter per sandwich • The reaction of nitrogen gas and hydrogen gas forming ammonia gas • Required: 1 molec. N2 gas and 3 molec. H2 gas 2 slices of br. +1 tbsp p.b. =1sndw. N2 NH3 NH3 N2 (g) + 3H2 (g) = 2 NH3 (g)

30. Limiting Reactant To determine the limiting reactant between two reactants: • Balance the equation • State the mole-mole relationships to make conversion factors • Convert the initial masses (reactants) to moles of each reactant • Calculate how many moles of product can be produced by each reactant

31. Limiting Reactant • Convert the moles of product to number of grams of product that each of the reactants would produce • Compare the numbers: The reactant producing the least amount of product (grams) is the limiting reactant

32. Limiting Reactant Problem • Lithium nitride, an ionic compound containing Li+ and N3- ions, is prepared by the reaction of lithium metal and nitrogen gas. Calculate the mass of lithium nitride formed from 56.0 g of nitrogen gas and 56.0 g of lithium metal. 2 6 X

33. Limiting Reactant Problem grams Li3N grams Li MM of Li MM of Li3N Stoichiometry moles Li3N moles Li moles Li3N moles Li 2 6 Find: g of Li3N Given: 56.0 g N2 Given: 56.0 g Li Equalities and Conversion Factors- 6 mol Li = 2 mol Li3N 1 mol N2 = 2 mol Li3N 1 mol Li = 6.941g Li 1 mol N2 = 28.00 g N2 1 mol Li3N= 34.83 g Li3N grams N2 Solution Map: MM of N2 moles N2 moles N2

34. Limiting Reactant Problem Limiting reactant 8.07 mol Li 93.67 g Li3N 2.00 mol N2 139.3 g Li3N

35. Limiting Reactant Problem • Lithium is the limiting reactant. We calculated the number of grams of lithium nitride which is formed in the reaction based on the limiting reactant • This is the calculated amount of lithium nitride formed if the reaction proceeds completely as described by its balanced chemical equation 93.7 g Li3N Theoretical yield

36. Percent Yield • The calculated amount of product that should be obtained is called the theoretical yield • Assumes all reactants are converted to product based on the mole-mole ratios of reactant to product • Rarely do you get the maximum amount of product • Side reactions • Loss during transfer • Accidental spills

37. Percent Yield • Theoretical Yield • The calculated amount of product • Actual Yield • The actual amount of product • Something less than the theoretical • Percent Yield • The fraction of the theoretical yield actually obtained is expressed as a percent

38. Percent Yield Example • Suppose, in the previous limiting reactant problem, you actually produced 90.8 g of Li3N. What is the percent yield of this reaction? 96.9 % yield

39. Enthalpy • Chemical reactions are associated with an absorption or evolution of heat • A change in energy occurs as bonds are broken (reactants) and new ones form (products) • Nearly all chemical reactions absorb or produce heat • Measured by the heat of reaction or enthalpy • Enthalpy change is the amount of heat produced or consumed in a process (∆H )

40. Sign of ∆Hrxn • Endothermic reactions absorb heat as they occur • If (∆H ) is positive, then heat is added to the reaction • If heat supply is removed, the reaction stops

41. Sign of ∆Hrxn • Exothermic reactions produce heat as they occur • If (∆H ) is negative, then heat is evolved by the reaction

42. Sign of ∆Hrxn Enthalpy of Reaction • Photosynthesis reaction • Carbon dioxide reacts with water to produce glucose and oxygen • Cell metabolism • Glucose reacts with oxygen to produce carbon dioxide and water ∆H = +2801 kJ ∆H = -2801 kJ

43. Stoichiometry of ∆Hrxn • The coefficients in a given chemical reaction represent the number of moles of reactants and products that produce the given heat of reaction (enthalpy change) • The combustion of methane gas: • This information gives a quantitative relationship between the heat evolved per mole of methane and oxygen gas ∆Hrxn = -890 kJ 2 mol O2 = -890 kJ 1 mol CH4 = -890 kJ

44. Stoichiometry of ∆Hrxn • If the combustion of 1 mol of CH4 with 2 mol O2 releases 890 kJ of heat, the combustion of 2 mol of CH4 with 4 mol of O2 produces twice as much heat • The equivalence statements can be used to make conversion factors between the amounts of reactants or products and the amount of heat absorbed or emitted in a given reaction • Calculate the amount of heat emitted when a certain amount in grams undergoes combustion ∆Hrxn = -1780 kJ

45. Stoichiometry Involving ∆H • Calculate the heat associated with the complete combustion of 4.50 g of methane gas ∆H rxn= -890 kJ Given: 4.50 g CH4 Find: kJ 1 mol CH4 = 16.00 g 1 mol CH4= -890 kJ Conversion Factors: Solution Map: g CH4 kJ CH4 mol CH4 -2.50 × 102 kJ Solution:

46. Stoichiometry Involving ∆H • The combustion of sulfur dioxide • It reacts with oxygen to produce sulfur trioxide • Calculate the heat produced when 75.2 g of sulfur trioxide is produced ∆H rxn= -99.1 kJ Given 75.2 g SO3 Find: Heat in kJ produced when SO3 is formed

47. Stoichiometry Involving ∆H Solution Map: Relation between g of SO3 and heat released Grams of SO3 Heat of rxn Moles of SO3 Molar mass kj Write the necessary conversion factors: Set up the problem: 46.5 kJ

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