1. Quantities in Chemical Reactions
Chapter 6:
Quantities in Chemical Reactions
Links for Chapter 6
Slide 29 Limit_React.exe, Limit_1.dir, Limit_2.dir, Limit_3.dir
Slide 41 chmvid12.mov NH4Cl + Ba(OH)2.8H2O
Slide 41 vd05_002.mov thermite
Slide 53 Exothermic system and surroundings.mov ????
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2. Questions for Consideration
What do the coefficients in balanced equations represent?
How can we use a balanced equation to relate the number of moles of reactants and products in a chemical reaction?
How can we use a balanced equation to relate the mass of reactants and products in a chemical reaction?
How do we determine which reactant limits the amount of product that can form?
How can we compare the amount of product we actually obtain to the amount we expect to obtain?
How can we describe and measure energy changes?
How are heat changes involved in chemical reactions?

3. Chapter 6 Topics: The Meaning of a Balanced Equation
Mole-Mole Conversions
Mass-Mass Conversions
Limiting Reactants
Percent Yield
Energy Changes
Heat Changes in Chemical Reactions

4. Introduction
How can we predict amounts of reactants and products in a reaction, such as that in an internal combustion engine?
How can we predict the amount of heat generated or absorbed during a reaction?
Figure 5.9

5. Internal Combustion Engine
In an ICE, octane burns with oxygen to produce hot gases that push against a piston to do work.
The amount of oxygen that reacts is dependent upon the amount of octane that burns.
The amount of energy produced also depends on the amount of octane that reacts.
Figure 6.3

6. 6.1 The Meaning of a Balanced Equation
What does a balanced equation tell us about the relative amounts of reactants and products? Let’s consider the combustion of propane:
Figure 6.4

7. What do the coefficients in a balanced chemical equation mean?
Balanced, the equation is:
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
The coefficients in the balanced equation tell us about the relative numbers of reactants that combine and products that form. There is an implied coefficient of 1 in front of C3H8(g).

8. What do the coefficients in a balanced chemical equation mean?
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
For every 1 molecule of propane that reacts with 5 molecules of oxygen gas, 3 molecules of carbon dioxide and 4 molecules of water are produced.
Figure 6.4

9. What do the coefficients in a balanced chemical equation mean?
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
The coefficients in the balanced equation also tell us about the relative numbers of moles of reactants and products. For every 1 mole of C3H8(g) that reacts with 5 moles of O2(g), 3 moles of CO2(g) and 4 moles of H2O(g) are formed.

10. Summary – The Meaning of the Coefficients

11. Amounts of Reactants and Products
The process of determining the amount of a reactant or product from another reactant or product in a reaction is called stoichiometry.

12. 6.2 Mole-Mole Conversions C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
A mole ratio is used to relate the number of moles of one reactant or product to another.
Mole ratios are obtained from the coefficients in the balanced equation.
For example, the mole ratio of O2 to C3H8 is 5:1 or:

13. Mole Ratios C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
The mole ratio of O2 to C3H8 allows us to calculate the amount of O2 that will react with any amount of C3H8 that reacts. The mole ratio is used as a conversion factor in a dimensional analysis equation.
If 0.40 mol C3H8 reacts:

14. Activity: Mole-Mole Conversions
Benzene (C6H6) burns in air according to the following equation:
2C6H6(g) + 15O2(g)  12CO2(g) + 6H2O(g)
What is the mole ratio of O2 to C6H6?
How many moles of O2 are required to react with each mole C6H6?
How many moles of O2 are required to react with 0.38 mole of C6H6?

15. Activity Solutions: Mole-Mole Conversions
2C6H6(g) + 15O2(g)  12CO2(g) + 6H2O(g)
What is the mole ratio of O2 to C6H6?
Based on the coefficients in the balanced chemical equation, the mole ratio can be written two ways:
15 moles O2 OR 2 moles C6H6
2 moles C6H moles O2
How many moles of O2 are required to react with each mole C6H6?
15/2 or 7.5 moles of O2 are required to react with each mole of C6H6
How many moles of O2 are required to react with 0.38 mole of C6H6?

16. Activity: Mole Ratios C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
What is the mole ratio for determining the moles of CO2 that will be produced when 2.3 mol O2 reacts?
How many moles of CO2 will be produced?

17. Activity: Mole Ratios 2Al(s) + 3Cl2(g)  2AlCl3(g)
How many moles of Cl2 are required to prepare 0.62 mol AlCl3?

18. 6.3 Mass-Mass Conversions
2Na(s) + Cl2(g)  2NaCl(s)
What mass of chlorine gas is required to react with 9.20 grams of sodium?
We don’t measure reactants and products in moles, but we commonly measure their mass.
The balanced equation does not tell us a mass relationship.
How do we convert grams of reactant or product to moles?

19. Mass-Mass Conversions
2Na(s) Cl2(g)  2NaCl(s)
(9.20 g Na)
1. We convert grams to moles using molar mass:
For a review, see Section 4.2
Figure from p. 129

20. Mass-Mass Conversions
2Na(s) Cl2(g)  2NaCl(s)
(9.20 g Na)
(0.400 mol)
2. Next we relate moles of Cl2 to moles of Na using the mole ratio:

21. Mass-Mass Conversions
2Na(s) Cl2(g)  2NaCl(s)
(9.20 g Na)
(0.400 mol Na) (0.200 mol Cl2)
3. The last step is to convert moles of Cl2 to grams using the molar mass of Cl2:

22. Mass-Mass Conversions
2Na(s) Cl2(g)  2NaCl(s)
(9.20 g Na) (14.2 g Cl2)
(0.400 mol Na) (0.200 mol Cl2)
3. The last step is to convert moles of Cl2 to grams using the molar mass of Cl2:

23. Steps for Mass-Mass Conversions
Figure 6.5

24. Mass-Mass Conversions
2Na(s) + Cl2(g)  2NaCl(s)
What mass of NaCl should be produced when 9.20 g Na reacts with 14.2 g Cl2?

25. Mass-Mass Conversions
2Na(s) + Cl2(g)  2NaCl(s)
What mass of NaCl should be produced when 9.20 g Na reacts with 14.2 g Cl2?
Figure 6.6

26. Activity: Mass-Mass Conversions
Given:
2 H2(g) + O2(g)  2 H2O(g)
If 1.8 × 108 g of hydrogen is used in the liftoff during a shuttle launch, then:
What mass of oxygen gas is consumed?
What mass of water vapor is produced?
Does the mass of the water vapor equal the masses of the reactants? If so, then what law describes this observation?

27. Activity Solutions: Mass-Mass Conversions
2 H2(g) + O2(g)  2 H2O(g)
What mass of oxygen gas is consumed?
Start with the number that was given in the problem –
1.8×108 g of H2(g). Convert to moles via the MM(H2), then
use a mole ratio to relate H2 to O2. Finally, convert from
moles of O2 to grams of O2 using the MM(O2).
Mass H2
(Given in
Problem)
Moles H2
Moles O2
Mass O2
MM(H2)
MM(O2)
Mole
Ratio

28. Activity Solutions: Mass-Mass Conversions
2 H2(g) + O2(g) → 2 H2O(g)
What mass of water vapor is produced? Start again with the number that was given in the problem: 1.8×108 of H2(g). Convert to moles using the MM(H2), then use a mole ratio to relate H2 to H2O. Finally, convert from moles of H2O to grams of H2O using the MM(H2O).
Mass H2
(Given in
Problem)
Moles H2
Moles
H2O
Mass H2O
MM(H2)
MM(H2O)
Mole
Ratio

29. Activity Solutions: Mass-Mass Conversions
Given:
2 H2(g) O2(g)  2 H2O(g)
1.8 × 108 g H × 109 g O2 = 1.6 × 109 g H2O
Does the mass of the water vapor equal the masses of the reactants?
Yes or close to it – hydrogen and oxygen added together equals 1.58 × 109 g H2O.
If so, then what law describes this observation?
The Law of Conservation of Mass

30. The Law of Conservation of Mass
Figure 6.6
The Law of Conservation of Mass states that the masses of the reactants that are consumed must equal the masses of the products that are formed.
Mass consumed reactants = Mass formed products

31. Activity: Mass-Mass Conversions
When aluminum metal is exposed to oxygen gas, a coating of aluminum oxide forms on the surface of the aluminum. The balanced equation for the reaction of aluminum metal with oxygen gas is:
4Al(s) + 3O2(g)  2Al2O3(s)
Suppose a sheet of pure aluminum gains g of mass when exposed to air. Assume that this gain can be attributed to its reaction with oxygen.
What mass of O2 reacted with the Al?
What mass of Al is reacted?
What mass of Al2O3 is formed?

32. Activity Solutions: Mass-Mass Conversions
4Al(s) + 3O2(g)  2Al2O3(s)
Suppose a sheet of pure aluminum gains g of mass when exposed to air. Assume that this gain can be attributed to its reaction with oxygen.
What mass of O2 reacted with the Al?
g O2
What mass of Al is reacted?
What mass of Al2O3 is formed?

33. 6.4 Limiting Reactants
When two reactants are mixed, one usually does not react completely because there is too much of it.
In this reaction solid magnesium metal reacts with aqueous hydrochloric acid (A). One reactant is in excess in B and the other reactant is in excess in C. Can you identify the excess reactant in each case?
Figure 6.7

34. Limiting Reactant
The limiting reactant is the reactant that reacts completely and is therefore not present when the reaction is complete.
Since the limiting reactant reacts completely, its amount determines the amount of product that can form.

35. What is the limiting reactant? Fe(s) + CuCl2(aq) 
Figure from
Example 6.3

36. Consider an Analogy: The construction of a model solar car
Figure 6.8

37. If you have 4 frames, 5 solar cells, 6 motors, and 12 wheels, how many solar cars can you make?
Figure 6.8

38. What is the limiting part?
Figure 6.8
The wheel is the limiting part because it will be completely used up.
The number of cars made depends on the number of wheels available. If there were 4 more wheels, then another car could have been made with the remaining parts.

39. How many solar cars?
Figure 6.8

40. Limiting Reactants at a Molecular Level
In a chemical reaction, the balanced equation tells us the relative number of molecules (or moles) that combine in the reaction.
Figure 6.9

41. Limiting Reactants at a Molecular Level
If reactants are not present in this ratio, then there will be a limiting reactant and excess of the other reactant.
Figure 6.10

42. Activity: Limiting Reactant 2H2(g) + O2(g)  2H2O(g)
Complete the after picture. How many H2O molecules form?
What is the limiting reactant? What is in excess?
Figure from p. 246

43. Activity Solution: Limiting Reactant
2H2(g) + O2(g)  2H2O(g)
What is the limiting reactant?
Using molecule ratios from the balanced chemical equation:
The limiting reactant is O2 because there is only enough of it to create 6 molecules of H2O.
Which reactant is left over?
H2 is left over at the end of the reaction.

44. 2H2(g) + O2(g)  2H2O(g)
You can also use “molecule” ratios to determine the limiting reactant. The 8 H2 molecules need 4 O2 molecules to react with them. There are only 3 O2 molecules, so all the H2 cannot react.
H2 is in excess, and O2 is the limiting reactant.
Figure from p. 246

45. Activity Solution: Limiting Reactant
The molecular-level diagram shows a mixture of reactant molecules (three O2 molecules and 8 H2 molecules) for the following reaction:
2H2(g) + O2(g)  2H2O(g)
The after picture should have 6 H2O molecules and 2 H2
molecules.
Figure from
p. 246

46. Steps for Determining the Limiting Reactant
Calculate the amount of one reactant (B) needed to react with the other reactant (A).
Compare the calculated amount of B (amount needed) to the actual amount of B that is given.
If calculated B = actual B, there is no limiting reactant. Both A and B will react completely.
If calculated B > actual B, B is the limiting reactant. Only B will react completely.
If calculated B < actual B, A is the limiting reactant. Only A will react completely.

47. Activity: Limiting Reactants (Mole Scale)
The balanced equation for the reaction of phosphorus and oxygen gas to form diphosphorus pentoxide is:
P4(s) + 5O2(g)  2P2O5(s)
What is the limiting reactant when each of the following sets of quantities of reactants is mixed?
0.50 mol P4 and 5.0 mol O2
0.20 mol P4 and 1.0 mol O2
0.25 mol P4 and 0.75 mol O2

48. Activity Solutions: Limiting Reactants (Mole Scale)
P4(s) + 5O2(g) → 2P2O5(s)
What is the limiting reactant when each of the following sets of quantities of reactants is mixed?
0.50 mol P4 and 5.0 mol O2
Each mole of P4 requires 5 moles of O2, so 0.50 mol P4 requires:
The amount of O2 present is more than the amount required, so P4 is the limiting reactant and O2 is present in excess.

49. Activity Solutions: Limiting Reactants (Mole Scale)
P4(s) + 5O2(g)  2P2O5(s)
What is the limiting reactant when each of the following sets of quantities of reactants is mixed?
0.20 mol P4 and 1.0 mol O2
Each mole of P4 requires 5 moles of O2, so 0.20 mol P4 requires:
Since this is the amount of O2 present, there is no limiting reactant. Both reactants are consumed completely.

50. Activity Solutions: Limiting Reactants (Mole Scale)
P4(s) + 5O2(g)  2P2O5(s)
What is the limiting reactant when each of the following sets of quantities of reactants is mixed?
0.25 mol P4 and 0.75 mol O2
Each mole of P4 requires 5 moles of O2, so 0.25 mol P4 requires:
Since the required amount of O2 is greater than that present, the limiting reactant is O2 and P4 is present in excess.

51. Limiting Reactants – Mole Scale
N2(g) + 3H2(g)  2NH3(g)
Identify the limiting reactant when the following are mixed:
2.0 mol N2 and 5.0 mole H2
3.10 mole N2 and 10.2 mol H2
How many moles of NH3 can be produced in each case?

52. Limiting Reactants – Mole Scale
N2(g) + 3H2(g) NH3(g)
Identify the limiting reactant when the following are mixed:
2.0 mol N2 and 5.0 mole H2
3.10 mole N2 and 10.2 mol H2
How many moles of NH3 can be produced in each case?
3.3 mole NH3
6.2 mole NH3

53. Limiting Reactants (Mass Scale)
Two approaches can be used when masses of the reactants are given. In each approach, it is necessary to convert masses of reactants to moles of reactants as a first step.
In the preferred approach, determine the limiting reactant on a mole scale as before, then proceed to calculate the mass.
In an alternate approach, calculate the amount of product predicted from each reactant. The calculation that gives the least amount of product identifies the limiting reactant.

54. Activity: Limiting Reactants (Mass Scale)
The balanced chemical equation for the reaction of aluminum metal and chlorine gas is:
2Al(s) + 3Cl2(g)  2AlCl3(s)
Assume 0.40 g Al is mixed with 0.60 g Cl2.
What is the limiting reactant?
What is the maximum amount of AlCl3, in grams, that can be produced?

55. Activity Solutions: Limiting Reactants (Mass Scale)
2Al(s) + 3Cl2(g)  2AlCl3(s)
Assume 0.40 g Al is mixed with 0.60 g Cl2.
What is the limiting reactant?
Calculate the moles of each reactant and determine which is the limiting reactant as in earlier examples:
For every 2 moles of Al, 3 moles of Cl2 are required, so moles of Al requires:
The limiting reactant is Cl2 because the required amount is greater than the amount present.

56. Activity Solutions: Limiting Reactants (Mass Scale)
2Al(s) + 3Cl2(g)  2AlCl3(s)
Assume 0.40 g Al is mixed with 0.60 g Cl2.
What is the maximum amount of AlCl3, in grams, that can be produced?
All of the mol Cl2 present in the system will react, so we use this amount to calculate the mass of product:

57. 6.5 Percent Yield
The amount of product actually obtained in the lab (actual yield) is usually less than the amount predicted by calculations (theoretical yield).
Yields describe the amount of product, and can be in mass units, moles, or number of molecules.
Figure 6.11

58. Percent Yield
The percent yield describes how much of a product is actually obtained relative to the amount that should form assuming complete reaction of the limiting reactant.

59. Activity: Percent Yield
2Na(s) + Cl2(g)  2NaCl(s)
If 0.20 mol chlorine reacts with excess sodium, and 0.40 mol NaCl are produced, what is the percent yield for the reaction?

60. Activity: Percent Yield
2Na(s) + Cl2(g)  2NaCl(s)
If mol chlorine reacts with excess sodium, and mol NaCl are produced, what is the percent yield for the reaction?

61. Activity: Percent Yield
A student was synthesizing aspirin in the laboratory. Using the amount of limiting reactant, she calculated the amount of aspirin that should form as 8.95 g. When she weighed her aspirin product on the balance, its mass was 7.44 g.
What is the actual yield of the aspirin?
What is the theoretical yield of the aspirin?
Calculate the percent yield for this synthesis.

62. Activity Solutions: Percent Yield
A student was synthesizing aspirin in the laboratory. Using the amount of limiting reactant, she calculated the amount of aspirin that should form as 8.95 g. When she weighed her aspirin product on the balance, its mass was 7.44 g.
What is the actual yield of the aspirin?
7.44 g aspirin
What is the theoretical yield of the aspirin?
8.95 g aspirin
Calculate the percent yield for this synthesis.

63. 6.6 Energy Changes
When methane reacts with oxygen when you use a gas stove, it’s obvious that an energy change is occurring. Heat is released to the surroundings and is used to heat water and cook food.
All chemical and physical changes are accompanied by energy changes.
Figure 5.8F

64. Law of Conservation of Energy
When energy is released from the atoms of a chemical reaction, it is transferred to the surroundings. The amount released by the reaction is equal to the amount absorbed by the surroundings.
Energy can also be converted from one type to another.
In an ICE, 32% of the energy is converted to work.
Figure 6.12

65. Conversion Efficiencies
Efficiency is the amount of useful work that is achieved from an energy conversion.
Figure 6.13

66. Efficiency
No conversion is totally efficient – some amount of energy is always lost to the surrounding as heat.
The efficiency of a conversion is the percentage that ends up in the form that we want.
Figure 6.14

67. Energy Changes that Accompany Chemical Reactions
Is energy absorbed or released by the reaction in each case? What happens to the temperature of the surroundings?
Figure 6.15

68. Endothermic and Exothermic Reactions
Most chemical reactions are exothermic – they release energy to the surroundings.
e.g. 2H2(g) + O2(g)  2H2O(g)
Some reactions are endothermic, absorbing energy from the surroundings and cooling them.
Figure 5.5
Figure from p. 244
Figure 5.8C

69. Quantities of Heat
Heat is energy that is transferred between two objects because of a difference in their temperatures.
In chemistry, quantities of energy (and heat) are usually expressed with units of joules (J) or calories (cal).
1 cal = J
Nutritionists use the Calorie (Cal) which is a kilocalorie (kcal), or 1000 cal.
1 Cal = 1000 cal = 1 kcal
Figure 6.16

70. Activity: Units of Energy
One serving of this cereal has 190 Calories.
What is the energy content in units of joules?
1 Cal = 1000 cal = 1 kcal
4.184 J = 1 cal
Figure 6.16

71. Specific Heat
When heat is added to a substance, the substance increases in temperature.
The amount of heat required to increase the temperature of 1 gram of a substance is dependent upon the identity of that substance, and is called the specific heat of that substance.
Specific heat units: J/(g ºC) or cal/(g ºC)
The specific heat of water is J/(g ºC). Other values are given in Table 6.2.

73. Activity: Specific Heat
How much heat is required to increase the temperature of 1.00 g of water by 10.0 ºC?
How much heat is required to increase the temperature of g of water by 1.00 ºC?

74. Activity: Example 6.11

75. Activity: Specific Heat
heat = mass × specific heat × temperature change
q = m × C × ΔT
How much heat must be added to 15.0 g of water to increase its temperature from 25.0 to 75.0 ºC?
What is the heat change when 15.0 grams of water cools from 75.0 to 25.0 ºC?

76. qsystem + qsurroundings = 0
Meaning of the Sign of q
qsystem + qsurroundings = 0
Figure 6.17

77. Activity: Specific Heat
The specific heat of aluminum is J/(g°C). If 156 g of aluminum at 75.0°C is cooled to 25.5°C, how much heat is transferred? What is the sign of q, and what is its significance?

78. Activity Solution: Specific Heat
The specific heat of aluminum is J/(g°C). If 156 g of aluminum at 75.0°C is cooled to 25.5°C, how much heat is transferred? What is the sign of q, and what is its significance?
q = mC∆T
∆T = Tf – Ti = 25.5°C – 75.0°C = -49.5°C
q = 156 g × J/(g°C) × -49.5°C
q = × 103 J
The sign of q is negative, which means that this process is exothermic and therefore releases 6.91 × 103 J into its surroundings.

79. Activity: Specific Heat
If 1 kJ of heat were added to each of the following 50.0-gram substances, which would increase in temperature by the greatest amount?
Aluminum: J/(g ºC)
Copper: J/(g ºC)
Lead: J/(g ºC)
The greatest temperature change would be for the metal with the smallest value of the specific heat, lead.

80. Calorimetry
Calorimetry is used to determine the heat change of a system by measuring the heat change of its surroundings.
In this calorimeter, an insulated cup is used so the surroundings is limited to what is inside the cup.
qsystem + qsurroundings = 0
Figure 6.18

81. Activity: Calorimetry
A 92.0-g piece of copper pipe is heated and then placed into this calorimeter with g of water at 25.00C. The final temperature of the mixture is 29.45C.
What is the heat change (q) of the pipe?
Figure 6.18

82. Activity Solution: Calorimetry
q = m × C × ΔT
qwater = g × J/(gºC) × (29.4525.00 ºC)
qwater = 1860 J
qpipe + qwater = 0
qpipe=  qwater
qpipe =  (1860 J)
qpipe =  1860 J
The copper pipe released 1860 J to the surroundings.
Figure 6.18
Figure 6.17

83. Activity: Calorimetry
A sample of a metal alloy is heated and then placed in g of water held in a calorimeter at 22.5°C. The final temperature of the water is 29.0°C. Assume heat exchange only occurs between the water and the alloy.
Was the initial temperature of the alloy greater than or less than the initial temperature of the water?
What is the heat change of the alloy?

84. Activity Solutions: Calorimetry
Was the initial temperature of the alloy greater than or less than the initial temperature of the water?
Since the temperature of the water increases, the initial temperature of the alloy must be greater than the initial temperature of the water.
What is the heat change of the alloy?
qalloy + qwater = 0
qalloy = -qwater
qwater = mwaterCwater∆Twater
∆Twater = Tf – Ti = 29.0°C – 22.5°C = 6.5°C
qwater = g × J/(g°C) × 6.5°C = 3.4 × 103 J
qalloy = -qwater = -3.4 × 103 J

85. 6.7 Heat Changes in Chemical Reactions
The heat change for a chemical reaction can be determined in the same way as the heat change for the piece of copper pipe, if the reaction takes place in solution.
qreaction+ qsurroundings = 0
If the reaction does not take place in solution, such as a combustion reaction, then a bomb calorimeter is required.

86. Bomb Calorimeter (For Non-Aqueous Reactions)
Figure 6.19

87. Bomb Calorimeter (For Non-Aqueous Reactions)
When an exothermic reaction takes place, the water in the calorimeter absorbs the heat from the reaction and increases in temperature.
qreaction = qcalorimeter
qreaction + qcalorimeter = 0
Figure 6.19

88. Activity: Heat Changes in Chemical Reactions
The balanced equation for the combination reaction of hydrogen gas and solid iodine is:
H2(g) + I2(s)  2HI(g)
The heat change q for this reaction is kJ per mole of I2 that reacts.
Is this reaction endothermic or exothermic?
What is the energy change when 2.50 mol of I2 reacts?

89. Activity Solutions: Heat Changes in Chemical Reactions
H2(g) + I2(s)  2HI(g)
The heat change q for this reaction is kJ per mole of I2 that reacts.
Is this reaction endothermic or exothermic?
Endothermic (indicated by the positive sign on q)
What is the energy change when 2.50 mol of I2 reacts?

90. Activity: Bomb Calorimeter
When 5.00 grams of methanol is burned in a calorimeter, the temperature of the calorimeter increases from 20.0 to 35.0ºC. The heat capacity of the calorimeter is 7.70 kJ/ ºC.
What is the heat of combustion per gram of methanol?
Figure 6.19

91. Activity Solution: Bomb Calorimeter
When 5.00 grams of methanol is burned in a calorimeter, the temperature of the calorimeter increases from 20.0 to 35.0ºC. The heat capacity of the calorimeter is 7.70 kJ/ ºC.
What is the heat of combustion per gram of methanol?
First calculate the heat change for the sample:
q = 7.70 kJ/oC × (35.0 – 20.0)oC = kJ
Then calculate the heat change per gram:
115.5 kJ/5.00 g = 23.1 kJ/g

92. Combustion of Fuels

93. Activity: Combustion of H2
The heat of combustion of H2 is -286 kJ/mol. How much heat is released when a balloon filled with mol H2 is burned?

94. Energy Content of Foods as presented in most nutritional labels
Figure 6.16