1. www.cengage.com/chemistry/cracolice Mark S. Cracolice Edward I. Peters Mark S. Cracolice The University of Montana Chapter 10 Quantity Relationships in Chemical Reactions

2. Conversion Factors from Equations The coefficients in a chemical equation give us the conversion factors to convert from the number of moles of one substance, to the number of moles of other substance in a reaction. 4 NH 3 (g) + 5 O 2 (g)  4 NO(g) + 6 H 2 O (g)

3. Conversion Factors from Equations Example: How many moles of water are formed by the reaction of 3.20 moles of ammonia NH 3 with oxygen. Step 1: Write and balance the chemical equation:

4. Conversion Factors from Equations 4 NH 3 (g) + 5 O 2 (g)  4 NO(g) + 6 H 2 O (g) Step2 Find what is given : 3.20 mol NH 3 Step 3 Find what is to be found: ? mol H 2 O Step 4 Find the path : mol NH 3  mol H 2 O Step 5 : Find the right conversion factor

5. Conversion Factors from Equations 4 NH 3 (g) + 5 O 2 (g)  4 NO(g) + 6 H 2 O (g) Step 6 Set up the calculation: 3.20 mol NH 3 x (6 mol H 2 O / 4 mol NH 3 ) = 4.80 mol H 2 O Note that 6 and 4 are exact numbers; they do not affect the significant figures in the final answer.

6. Conversion Factors from Equations Example 2 How many moles of oxygen are required to burn 2.40 moles of ethane, C 2 H 6 ? Step 1: Write the chemical equation 2 C 2 H 6 (g) + 7 O 2 (g)  4 CO 2 (g) + 6 H 2 O (l) conversionconversion factor Number of moles C 2 H 6  number of moles O 2 : 7 mol O 2 / 2 mol C 2 H 6 Number of moles C 2 H 6  number of moles H 2 O : 6 mol H 2 O / 2 mol C 2 H 6 Number of moles C 2 H 6  number of moles CO 2 : 4 mol CO 2 / 2 mol C 2 H 6 Number of moles O 2  number of moles CO 2 : 4 mol CO 2 / 7 mol O 2 Number of moles O 2  number of moles H 2 O : 6 mol H 2 O / 7 mol O 2 Number of moles CO 2  number of moles H 2 O 6 mol H 2 O / 4 mol CO 2 :

7. Conversion Factors from Equations 2 C 2 H 6 (g) + 7 O 2 (g)  4 CO 2 (g) + 6 H 2 O (l) Step2 Find what is given : 2.40 mol C 2 H 6 Step 3 Find what is to be found: ? mol O 2 Step 4 Find the path : mol C 2 H 6  mol O 2 Step 5 : Find the right conversion factor 7 mol O 2 / 2 mol C 2 H 6

8. Conversion Factors from Equations 2 C 2 H 6 (g) + 7 O 2 (g)  4 CO 2 (g) + 6 H 2 O (l) Step 6 Set up the calculation: 2.40 mol C 2 H 6 x (7 mol O 2 / 2 mol C 2 H 6 ) = 8.40 mol O 2 Note that 7 and 2 are exact numbers; they do not affect the significant figures in the final answer.

9. Conversion Factors from Equations Example 3 Ammonia is formed from its elements. How many moles of hydrogen are needed to produce 4.20 moles NH 3. Chemical equation : N 2 + 3 H 2  2 NH 3

10. Conversion Factors from Equations N 2 + 3 H 2  2 NH 3 Given 4.20 moles NH 3 Wanted mol H 2 Path 4.20 moles NH 3  mol H 2 Factor 3 mol H 2 / 2 moles NH 3 Calculation 4.20 moles NH 3 x (3 mol H 2 / 2 moles NH 3 ) = 6.30 mol NH 3

11. Mass–Mass Stoichiometry How to Solve a Stoichiometry Problem: The Stoichiometry Path Step 1: Change the mass of the given species to moles. Step 2: Change the moles of the given species to the moles of the wanted species. Step 3: Change the moles of the wanted species to mass.

12. Mass–Mass Stoichiometry Mass-to-Mass Stoichiometry Path Mass ofMoles of Moles ofMass of GivenGiven WantedWanted Molar Mass Equation Molar Mass coefficients Mass Given × × ×

13. Mass–Mass Stoichiometry Example 1 Calculate the number of grams of carbon dioxide produced by burning 66.0 g of heptanes C 7 H 16 (l) Step 1 Chemical reaction : C 7 H 16 (l) + 11 O 2 (g)  7 CO 2 (g) + 8 H 2 O (l) Step2 Find what is given : 66.0 g C 7 H 16 Step 3 Find what is to be found: ? g of CO 2

14. Mass–Mass Stoichiometry C 7 H 16 (l) + 11 O 2 (g)  7 CO 2 (g) + 8 H 2 O (l) Step 4 Find the path : g of C 7 H 16  mol C 7 H 16  mol CO 2  g of CO 2 Step 5 : Find the right conversion factors g of C 7 H 16  mol C 7 H 16 : 1 mol C 2 H 6 / 100.20 g C 7 H 16 mol C 7 H 16  mol CO 2 7 mol CO 2 / 1 mol C 7 H 16 mol CO 2  g of CO 2 44.01 g CO 2 / 1 mol CO 2

15. Mass–Mass Stoichiometry C 7 H 16 (l) + 11 O 2 (g)  7 CO 2 (g) + 8 H 2 O (l) Step 6 Set up the calculation: 66.0 g C 7 H 16 x (1 mol C 7 H 16 / 100.20 g C 7 H 16 ) x (7 mol CO 2 / 1 mol C 7 H 16 ) x (44.01 g CO 2 / 1mol CO 2 ) = 203 g of CO 2

16. Percent Yield The actual yield of a chemical reaction is usually less than the ideal yield predicted by a stoichiometry calculation because: reactants may be impure the reaction may not go to completion other reactions may occur Actual yield is experimentally determined.

17. Percent Yield Percent yield expresses the ratio of actual yield to ideal yield: % yield = × 100%

18. Percent Yield Example Calculate the theoretical yield of carbon dioxide and the percent yield when burning of 66.0 grams of C 7 H 16 produced 181 grams of CO 2. C 7 H 16 (l) + 11 O 2 (g)  7 CO 2 (g) + 8 H 2 O (l) Theoretical yield of CO 2 66.0 g x ( 1mol C 7 H 16 / 100.20 g C 7 H 16 ) x ( 7 mol CO 2 / 1 mol. C 7 H 16 ) x ( 44.01 CO 2 / 1 mol. CO 2 ) = = 203 g CO 2 Percent yield = ( 181g/203g) x 100 = 89.2 %

19. Percent Yield Example: Consider the reaction of hydrogen and nitrogen that forms ammonia with a 92.2% yield. What quantity of ammonia will be produced by reacting 125 g of hydrogen with excess nitrogen? Solution: Use 92.2% yield as a P ER expression: 3 H 2 + N 2 2 NH 3 125 g H 2 × × × × = 649 g NH 3

20. Limiting Reactants Three atoms of carbon react with two molecules of oxygen: C(g) + O 2 (g) CO 2 (g)

21. Limiting Reactants The reaction will stop when two molecules of oxygen are used up. Two carbon dioxide molecules will form; One carbon atom will remain unreacted: C + O 2 CO 2 Start 3 2 0 Used (+) or Produced (–)– 2– 2+ 2 Finish 1 0 2

22. Limiting Reactants Limiting Reactant The reactant that is completely used up. (Oxygen) Excess Reactant The reactant initially present in excess. (some will remain unreacted) (Carbon)

23. Limiting Reactants The reactant that is completely used up by the reaction, is called the limiting reactant. Other reactants have some excess which will remain unreacted. There are two approaches to solving limiting reactant problems: the comparison of moles method and the smaller amount method.

24. Limiting Reactants: Compare Moles Comparison-of-Moles Method How to Solve a Limiting Reactant Problem: 1. Convert the number of grams of each reactant to moles. 2. Identify the limiting reactant by comparing the theoretical mole ratio of reactants to the actual mole ratio. 3. Calculate the number of moles of each species that reacts or is produced using the limiting reactant number of moles. 4. Calculate the number of moles of each species that remains after the reaction. 5. Change the number of moles of each species to grams.

25. Limiting Reactants: Compare Moles Example Calculate the mass of antimony(III) iodide that can be produced by the reaction of 129 g antimony, Sb (Z=51), and 381 g iodine. Also find the number of grams of the element that will be left. 2 Sb + 3 I 2  2 SbI 3 Calculate the number of moles: Moles of Sb = 129 g Sb x (1 mol Sb/ 121.8 g Sb) = 1.06 mol Sb Moles of I 2 = 381 g I 2 x (1 mol I 2 / 253.8 g I 2 ) = 1.50 mol I 2

26. Limiting Reactants: Compare Moles 2 Sb + 3 I 2  2 SbI 3 Theoretical mole ratio: mol of I 2 / mol of Sb = 3 mol of I 2 / 2 mol of Sb = 1.5 mol of I 2 / 1 mol of Sb Actual mole ratio: mol of I 2 / mol of Sb = 1.5 mol of I 2 / 1.06 mol of Sb =1.42 mol of I 2 / 1 mol of Sb Actual mole ratio of iodine over antimony is smaller than theoretical mole ratio, so iodine is the limiting reactant and antimony is in excess.

27. Limiting Reactants: Compare Moles 2 Sb + 3 I 2  2 SbI 3 The number of mole Sb used = 1.50 mol of I 2 x ( 2 mol of Sb/ 3 mol of I 2 ) = 1.00 mol of Sb mol of Sb in excess = (1.06 - 1.00) mol of Sb = 0.06 mol of Sb mass of Sb in excess = 0.06 mol of Sb x (121.8 g Sb/mol Sb) = 7 g Sb

28. Limiting Reactants: Compare Moles 2 Sb + 3 I 2  2 SbI 3 The number of mole SbI 3 produced = 1.50 mol of I 2 x ( 2 mol of SbI 3 / 3 mol of I 2 ) = 1.00 mol of SbI 3 mass of SbI 3 produced = 1.00 mol of SbI 3 x (502.5 g SbI 3 /mol SbI 3 ) = 502.5 g SbI 3

29. Limiting Reactants: Smaller Amount How to Solve a Limiting Reactant Problem by Smaller Amount Method 1. Calculate the amount of product that can be formed by the initial amount of each reactant. a)The reactant that yields the smaller amount of product is the limiting reactant. b)The smaller amount of product is the amount that will be formed when all of the limiting reactant is used up. 2. Calculate the amount of excess reactant that is used by the total amount of limiting reactant. 3. Subtract from the amount of excess reactant present initially the amount that is used by all of the limiting reactant. The difference is the amount of excess reactant that is left.

30. Limiting Reactants: Smaller Amount Example Calculate the mass of antimony(III) iodide that can be produced by the reaction of 129 g antimony, Sb (Z=51), and 381 g iodine. Also find the number of grams of the element that will be left. 2 Sb + 3 I 2  2 SbI 3 Solution: Step 1 is to calculate the amount of product that can be formed by the initial amount of each reactant.

31. Limiting Reactants: Smaller Amount 2 Sb + 3 I 2  2 SbI 3 First, assume that antimony is the limiting reactant. Grams of SbI 3 produced = 0.129g Sb x(1 mol Sb/121.8 gSb) x (2mol SbI 3 /2mol Sb) x (502.5 g SbI 3 /molSbI 3 ) = = 532 g SbI 3

32. Limiting Reactants: Smaller Amount 2 Sb + 3 I 2  2 SbI 3 Second, assume that iodine is the limiting reactant. grams of SbCl 3 produced = 381g I 2 x ( mol I 2 / 253.8 g I 2 ) x (2mol SbI 3 /3mol I 2 ) x (502.5 g SbI 3 / mol SbI 3 ) = = 503 g SbI 3 The reactant iodine, which yields the smaller amount (503 g) of SbI 3, is the limiting reactant. The amount of antimony(III) iodide produced is 503 g.

33. Limiting Reactants: Smaller Amount 2 Sb + 3 I 2  2 SbI 3 The amount of antimony required is calculated as follows Grams of antimony required = 381 g I 2 x (1 mol I 2 /253.8 g I 2 ) x (2 mol Sb/ 3 mol I 2 ) x (121.8 g Sb/mol Sb) = 122 g Sb The amount of antimony in excess = 129 g (initial) – 122g (used) = 7 g Sb (left)

34. Energy The ability to do work or transfer heat. SI (derived) energy unit: Joule Joule (J): The amount of energy exerted when a force of one newton (force required to cause a mass of 1 kg to accelerate at a rate of 1 m/s 2 ) is applied over a displacement of one meter: 1 joule = 1 newton × 1 meter 1 J = 1 kg m 2 /sec 2

35. Energy Non-SI metric energy unit: Calorie calorie (historical definition): Amount of energy required to raise the temperature of 1 g of water by 1°C. calorie (modern definition) 1 cal 4.184 J A food Calorie (Cal) is a thermochemical kilocalorie. In scientific writing it is capitalized; in everyday writing, interpret the context. 1 kcal = 4.184 kJ

36. Thermochemical Equations Enthalpy H = E + PV Where E is the internal energy of the system, P is the pressure of the system, and V is the volume of the system. P, V, E, H which depend only on the present state of the system, and do not depend on the system’s past or future, are called state functions. ∆H is the enthalpy of reaction. ∆ means “change in”: It is determined by subtracting the initial quantity from the final quantity

37. Thermochemical Equations Enthalpy of Reaction, ∆H It can be demonstrated that for a reaction studied at constant pressure the heat of reaction is a measure of the change in enthalpy 1 for the system. Heat of reaction = H(products) - H (reactants ) = Δ H When a system gives off heat (reaction is exothermic) enthalpy of the system goes down and Δ H has a negative value. When a reaction absorbs heat (reaction is endothermic) enthalpy increases and Δ H is positive..

38. Thermochemical Equations There are two ways to write a thermochemical Equation Method 1 The ΔH of the reaction is written to the right of the equation 2 C 2 H 6 (g) + 7O 2 (g)  4CO 2 (g) + 6H 2 O (g) ΔH = -2855 kJ Method 2 Energy is included in the thermochemical equation as if it were a reactant or product. 2 C 2 H 6 (g) + 7O 2 (g)  4CO 2 (g) + 6H 2 O (g) + 2855 kJ

39. Thermochemical Stoichiometry Since there is a proportional relationship between moles of different substances and heat of reaction, conversion factors can be written between kilojoules and moles of any substance. These factors are used in solving thermochemical stoichiometry problems.

40. Thermochemical Stoichiometry Example : How many kilojoules of heat are released when 73.0 g C 2 H 6 (g) burn Equation : 2 C 2 H 6 (g) + 7O 2 (g)  4CO 2 (g) + 6H 2 O (l) + 3119 kJ

41. Thermochemical Stoichiometry 2 C 2 H 6 (g) + 7O 2 (g)  4CO 2 (g) + 6H 2 O (l) + 3119 kJ Given 73.0 g C 2 H 6 (g) Wanted : kJ Path g C 2 H 6 (g)  mol C 2 H 6 (g)  kJ Factors (1 mol C 2 H 6 / 30.07 g C 2 H 6 ) (3119 kJ/ 2 mol C 2 H 6 (g))

42. Thermochemical Stoichiometry 2 C 2 H 6 (g) + 7O 2 (g)  4CO 2 (g) + 6H 2 O (l) + 3119 kJ Solve the problem: 73.0 g C 2 H 6 (g) x (1 mol C 2 H 6 / 30.07 g C 2 H 6 ) x (3119 kJ/ 2 mol C 2 H 6 (l) ) = = 3.79 x 10 3 kJ

43. Thermochemical Stoichiometry Example: When propane, C 3 H 8 (g), is burned to form gaseous carbon dioxide and liquid water, 2.22 × 10 3 kJ of heat is released for every mole of propane burned. What quantity of heat is released when 1.00 × 10 2 g of propane is burned? Solution: First, write and balance the thermochemical equation to determine the stoichiometric relationships. C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l) ∆H = –2.22x10 3 kJ

44. Thermochemical Stoichiometry C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l) ∆H = –2.22 × 10 3 kJ G IVEN: 1.00 × 10 2 g C 3 H 8 1 mol C 3 H 8 /44.09 g C 3 H 8 P ATH: g C 3 H 8 mol C 3 H 8 2.22 × 10 3 kJ/1 mol C 3 H 8 kJ 1.00 × 10 2 g C 3 H 8 × × = 5.04 × 10 3 kJ

45. HOMEWORK Homework: 1, 7, 23, 35, 45, 47, 67.