1. 7.10 Classifying Chemical Reactions by What Atoms Do

2. Quantities in Chemical Reactions
Chapter 8
Quantities in Chemical Reactions

3. 8.2 Stoichiometry: Relationships between Ingredients
The numerical relationship between chemical quantities in a balanced chemical equation is called reaction stoichiometry.
We can predict the amounts of products that form in a chemical reaction based on the amounts of reactants.
We can predict how much of the reactants are necessary to form a given amount of product.
We can predict how much of one reactant is required to completely react with another reactant.

4. 8.2 Making Pancakes: Relationships between Ingredients
A recipe gives numerical relationships between the ingredients and the number of pancakes.

5. 8.2 Making Pancakes: Relationships between Ingredients
The recipe shows the numerical relationships between the pancake ingredients.
If we have 2 eggs—and enough of everything else—we can make 5 pancakes.
We can write this relationship as a ratio.
2 eggs:5 pancakes

6. What if we have 8 eggs? Assuming that we have enough of everything else, how many pancakes can we make?

7. 8.3 Making Molecules: Mole-to-Mole Conversions
In a balanced chemical equation, we have a “recipe” for how reactants combine to form products.
The following equation shows how hydrogen and nitrogen combine to form ammonia (NH3).

8. 3 H2 molecules : 1 N2 molecule : 2 NH3 molecules
3 H2(g) + N2(g)  2 NH3(g)
The balanced equation shows that 3 H2 molecules react with 1 N2 molecule to form 2 NH3 molecules.
We can express these relationships as ratios.
3 H2 molecules : 1 N2 molecule : 2 NH3 molecules
Since we do not ordinarily deal with individual molecules, we can express the same ratios in moles.
3 mol H2 : 1 mol N2 : 2 mol NH3

9. 3 H2(g) + N2(g)  2 NH3(g)
If we have 3 mol of N2, and more than enough H2, how much NH3 can we make?

10. Example 8.1 Mole-to-Mole Conversions
Sodium chloride, NaCl, forms by this reaction between sodium and chlorine.
2 Na(s) + Cl2(g)2 NaCl(s)
How many moles of NaCl result from the complete reaction of 3.4 mol of Cl2? Assume that there is more than
enough Na.
SORT
You are given the number of moles of a reactant (Cl2) and asked to find the number of moles of product (NaCl) that will form if the reactant completely reacts.
STRATEGIZE
Draw the solution map beginning with moles of chlorine and using the stoichiometric conversion factor to calculate moles of sodium chloride. The conversion factor comes from the balanced chemical equation.
GIVEN: 3.4 mol Cl2
FIND: mol NaCl
SOLUTION MAP
RELATIONSHIPS USED
1 mol Cl2 : 2 mol NaCl (from balanced chemical equation)

11. Example 8.1 Mole-to-Mole Conversions
Continued
SOLVE
Follow the solution map to solve the problem. There is enough Cl2 to produce 6.8 mol of NaCl.
SOLUTION
CHECK
Check your answer. Are the units correct? Does the answer make physical sense?
The answer has the correct units, moles. The answer is reasonable because each mole of Cl2 makes two moles of NaCl.
SKILLBUILDER 8.1 Mole-to-Mole Conversions
Water is formed when hydrogen gas reacts explosively with oxygen gas according to the balanced equation:
O2(g) + 2 H2(g)2 H2O(g)
How many moles of H2O result from the complete reaction of 24.6 mol of O2? Assume that there is more than enough H2.
Answer: 49.2 mol H20
For More Practice Example 8.8; Problems 15, 16, 17, 18.

12. 8.4 Making Molecules: Mass-to-Mass Conversions
A chemical equation contains conversion factors between moles of reactants and moles of products.
We are often interested in relationships between mass of reactants and mass of products.
The general outline for this type of calculation is:

13. 2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
What mass of carbon dioxide is emitted by an automobile per 5.0 x 102 g pure octane used?
The balanced chemical equation gives us a relationship between moles of C8H18 and moles of CO2.
Before using that relationship, we must convert from grams to moles.

14. 2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
SOLUTION:

15. 2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
SOLUTION:

16. Example 8.2 Mass-to-Mass Conversions
In photosynthesis, plants convert carbon dioxide and water into glucose (C6H12O6) according to the reaction:
How many grams of glucose can be synthesized from 58.5 g of CO2? Assume that there is more than enough water present to react with all of the CO2.
SORT
You are given the mass of carbon dioxide and asked to find the mass of glucose that can form if the carbon dioxide completely reacts.
STRATEGIZE
The solution map uses the general outline
Mass A  Moles A 
Moles B  Mass B
where A is carbon dioxide and B is glucose.
GIVEN: g CO2
FIND: g C6H12O6
SOLUTION MAP

17. Example 8.2 Mass-to-Mass Conversions
Continued
The main conversion factor is the stoichiometric relationship between moles of carbon dioxide and moles of glucose. This conversion factor comes from the balanced equation. The other conversion factors are simply the molar masses of carbon dioxide and glucose.
SOLVE
Follow the solution map to solve the problem. Begin with grams of carbon dioxide and multiply by the appropriate factors to arrive at grams of glucose.
RELATIONSHIPS USED
6 mol CO2 : 1 mol C6H12O6 (from balanced chemical equation)
Molar mass CO2 = g/mol
Molar mass C6H12O6 = g/mol
SOLUTION
CHECK
Are the units correct? Does the answer make physical sense?
The units, g C6H12O6, are correct. The magnitude of the answer seems reasonable because it is of the same order of magnitude as the given mass of carbon dioxide. An answer that is orders of magnitude different would immediately be suspect.

18. Mg(OH)2(aq) + 2 HCl(aq)  2 H2O(l) + MgCl2(aq)
Example 8.2 Mass-to-Mass Conversions
Continued
SKILLBUILDER 8.2 Mass-to-Mass Conversions
Magnesium hydroxide, the active ingredient in milk of magnesia, neutralizes stomach acid, primarily HCl, according to the reaction:
Mg(OH)2(aq) + 2 HCl(aq)  2 H2O(l) + MgCl2(aq)
How much HCl in grams can be neutralized by 5.50 g of Mg(OH)2?
Answer: 6.88 g HCl
For More Practice Example 8.9; Problems 31, 32, 33, 34.

19. 4 NO2(g) + O2(g) + 2 H2O(l)  4 HNO3(aq)
Example 8.3 Mass-to-Mass Conversions
One of the components of acid rain is nitric acid, which forms when NO2, a pollutant, reacts with oxygen and rainwater according to the following simplified reaction.
4 NO2(g) + O2(g) + 2 H2O(l)  4 HNO3(aq)
Assuming that there is more than enough O2 and H2O, how much HNO3 in kilograms forms from 1.5  103 kg of NO2 pollutant?
SORT
You are given the mass of nitrogen dioxide (a reactant) and asked to find the mass of nitric acid that can form if the nitrogen dioxide completely reacts.
GIVEN:  103 kg NO2
FIND: kg HNO3

20. Example 8.3 Mass-to-Mass Conversions
Continued
STRATEGIZE
The solution map follows the general format of:
Mass  Moles 
Moles  Mass
However, since the original quantity of NO2 is given in kilograms, you must first convert to grams. Since the final quantity is requested in kilograms, you must convert back to kilograms at the end. The main conversion factor is the stoichiometric relationship between moles of nitrogen dioxide and moles of nitric acid. This conversion factor comes from the balanced equation. The other conversion factors are simply the molar masses of nitrogen dioxide and nitric acid and the relationship between kilograms and grams.
SOLUTION MAP
RELATIONSHIPS USED
4 mol NO2 : 4 mol HNO3 (from balanced chemical equation)
Molar mass NO2 = g/mol
Molar mass HNO3 = g/mol
1 kg = 1000 g

21. 2 SO2(g) + O2(g) + 2 H2O(l)  2 H2SO4(aq)
Example 8.3 Mass-to-Mass Conversions
Continued
SOLVE
Follow the solution map to solve the problem. Begin with kilograms of nitrogen dioxide and multiply by the appropriate conversion factors to arrive at kilograms of nitric acid.
SOLUTION
CHECK
Are the units correct? Does the answer make physical sense?
The units, kg HNO3 are correct. The magnitude of the answer seems reasonable because it is of the same order of magnitude as the given mass of nitrogen dioxide. An answer that is orders of magnitude different would immediately be suspect.
SKILLBUILDER 8.3 Mass-to-Mass Conversions
Another component of acid rain is sulfuric acid, which forms when SO2, also a pollutant, reacts with oxygen and rainwater according to the following reaction.
2 SO2(g) + O2(g) + 2 H2O(l)  2 H2SO4(aq)
Assuming that there is more than enough O2 and H2O, how much in kilograms forms from H2SO4 2.6  103 kg of SO2?
Answer: 4.0  103 kg H2SO4
For More Practice Problems 35, 36, 37, 38.

22. Administrative Stuff
Silence your electronic devices and make sure your clickers are ready.
Chapter 7 Homework is due TODAY at 5pm.
Continue learning the polyatomic ions and solubility rules from chapters 5 and 6.
Exam 2 is 10/14/13 and covers Chapters 6, 7, 8, and 9.
We have 5 classes between now and then including today.

23. 8.5 More Pancakes: Limiting Reactant, Theoretical Yield, and Percent Yield
Suppose we have 3 cups flour, 10 eggs, and 4 tsp baking powder.
How many pancakes can we make?
We have enough flour for 15 pancakes, enough eggs for 25 pancakes, and enough baking powder for 40 pancakes.

24. 8.5 More Pancakes: Limiting Reactant, Theoretical Yield, and Percent Yield
If this were a chemical reaction, the flour would be the limiting reactant and 15 pancakes would be the theoretical yield.

25. 8.5 More Pancakes: Limiting Reactant, Theoretical Yield, and Percent Yield
Suppose we cook our pancakes. We accidentally burn three of them and one falls on the floor.
So even though we had enough flour for 15 pancakes, we finished with only 11 pancakes.
If this were a chemical reaction, the 11 pancakes would be our actual yield, the amount of product actually produced by a chemical reaction.

26. Since four of the pancakes were ruined,
8.5 More Pancakes: Limiting Reactant, Theoretical Yield, and Percent Yield
Our percent yield, the percentage of the theoretical yield that was actually attained, is:
Since four of the pancakes were ruined,
we got only 73% of our theoretical yield.

27. Actual Yield and Percent Yield
The actual yield of a chemical reaction must be determined experimentally and depends on the reaction conditions.
The actual yield is almost always less than 100%.
Some of the product does not form.
Product is lost in the process of recovering it.

28. Limiting Reactant, Theoretical Yield, Actual Yield, and Percent Yield
To summarize: • Limiting reactant (or limiting reagent)—the reactant that is completely consumed in a chemical reaction. • Theoretical yield—the amount of product that can be made in a chemical reaction based on the amount of limiting reactant. • Actual yield—the amount of product actually produced by a chemical reaction. • Percent yield—(actual yield/theoretical yield)*100%.

29. Example: Ti(s) + 2 Cl2(g)  TiCl4(s) Given (moles): 1. 8 mol Ti and 3
Example: Ti(s) + 2 Cl2(g)  TiCl4(s) Given (moles): 1.8 mol Ti and 3.2 mol Cl2 Find: limiting reactant and theoretical yield SOLUTION MAP:

30. Example: Ti(s) + 2 Cl2(g)  TiCl4(s) Given (moles): 1. 8 mol Ti and 3
Example: Ti(s) + 2 Cl2(g)  TiCl4(s) Given (moles): 1.8 mol Ti and 3.2 mol Cl2 Find: limiting reactant and theoretical yield SOLUTION:

31. Limiting Reactant, Theoretical Yield, Actual Yield, and Percent Yield
In many industrial applications, the more costly reactant or the reactant that is most difficult to remove from the product mixture is chosen to be the limiting reactant.
When working in the laboratory, we measure the amounts of reactants in grams.
To find limiting reactants and theoretical yields from initial masses, we must add two steps to our calculations.

32. Example: Na(s) + Cl2(g)  2 NaCl(s) Given (grams): 53. 2 g Na and 65
Example: Na(s) + Cl2(g)  2 NaCl(s) Given (grams): 53.2 g Na and 65.8 g Cl2 Find: limiting reactant and theoretical yield SOLUTION MAP:

33. Example: Na(s) + Cl2(g)  2 NaCl(s) Given (grams): 53. 2 g Na and 65
Example: Na(s) + Cl2(g)  2 NaCl(s) Given (grams): 53.2 g Na and 65.8 g Cl2 Find: limiting reactant and theoretical yield SOLUTION:

34. Example: Na(s) + Cl2(g)  2 NaCl(s) Given (grams): actual yield 86
Example: Na(s) + Cl2(g)  2 NaCl(s) Given (grams): actual yield 86.4 g NaCl Find: percent yield
The actual yield is usually less than the theoretical yield because at least a small amount of product is lost or does not form during a reaction.

35. Cu2O(s) + C(s)  2 Cu(s) + CO(g)
Example 8.6 Finding Limiting Reactant, Theoretical Yield, and Percent Yield
Consider the reaction:
Cu2O(s) + C(s)  2 Cu(s) + CO(g)
When 11.5 g of C are allowed to react with g of Cu2O, 87.4 g of Cu are obtained. Find the limiting reactant, theoretical yield, and percent yield.
SORT
You are given the mass of the reactants, carbon and copper(I) oxide, as well as the mass of copper formed by the reaction. You are asked to find the limiting reactant, theoretical yield, and percent yield.
GIVEN: g C
114.5 g Cu2O
87.4 g Cu produced
FIND: limiting reactant
theoretical yield
percent yield

36. Example 8.6 Finding Limiting Reactant, Theoretical Yield, and Percent Yield
Continued
STRATEGIZE
The solution map shows how to find the mass of Cu formed by the initial masses of Cu2O and C. The reactant that makes the least amount of product is the limiting reactant and determines the theoretical yield.
The main conversion factors are the stoichiometric relationships between moles of each reactant and moles of copper. The other conversion factors are the molar masses of copper(I) oxide, carbon, and copper.
SOLUTION MAP
RELATIONSHIPS USED
1 mol Cu2O: 2 mol Cu
1 mol C : 2 mol Cu
Molar mass Cu = g/mol
Molar mass C = g/mol
Molar mass Cu2O = g/mol

37. Example 8.6 Finding Limiting Reactant, Theoretical Yield, and Percent Yield
Continued
SOLVE
Follow the solution map, beginning with the actual amount of each reactant given, to calculate the amount of product that can be made from each reactant.
Since Cu2O makes the least amount of product, Cu2O is the limiting reactant. The theoretical yield is then the amount of product made by the limiting reactant. The percent yield is the actual yield (87.4 g Cu) divided by the theoretical yield (101.7 g Cu) multiplied by 100%.
CHECK
Are the units correct? Does the answer make physical sense?
SOLUTION
The theoretical yield has the right units (g Cu). The magnitude of the theoretical yield seems reasonable because it is of the same order of magnitude as the given masses of C and Cu2O. The theoretical yield is reasonable because it is less than 100%. Any calculated theoretical yield above 100% would be suspect.

38. Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g)
Example 8.6 Finding Limiting Reactant, Theoretical Yield, and Percent Yield
Continued
SKILLBUILDER 8.6 Finding Limiting Reactant, Theoretical Yield, and Percent Yield
This reaction is used to obtain iron from iron ore:
Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g)
The reaction of 185 g of Fe2O3 with 95.3 g of CO produces 87.4 g of Fe. Find the limiting reactant, theoretical yield, and percent yield.
Answer: Limiting reactant is CO;
theoretical yield = 127 g Fe;
percent yield = 68.8%
For More Practice Example 8.10; Problems 61, 62, 63, 64, 65, 66.

39. 8.7 Enthalpy: A Measure of the Heat Evolved or Absorbed in a Reaction
Chemical reactions can be exothermic (they emit thermal energy when they occur).
Chemical reactions can be endothermic (they absorb thermal energy when they occur).
The amount of thermal energy emitted or absorbed by a chemical reaction, under conditions of constant pressure (which are common for most everyday reactions), can be quantified with a function called enthalpy.

40. 8.7 Enthalpy: A Measure of the Heat Evolved or Absorbed in a Reaction
We define the enthalpy of reaction, ΔHrxn, as the amount of thermal energy (or heat) that flows when a reaction occurs at constant pressure.
The sign of ΔHrxn (positive or negative) depends on the direction in which thermal energy flows when the reaction occurs.
Energy flowing out of the chemical system is like a withdrawal and carries a negative sign.
Energy flowing into the system is like a deposit and carries a positive sign.

41. Figure 8.3 Exothermic and Endothermic reactions
(a) In an exothermic reaction, energy is released into the surroundings. (b) In an endothermic reaction, energy is absorbed from the surroundings.

42. Sign of ΔHrxn
When thermal energy flows out of the reaction and into the surroundings (as in an exothermic reaction), then ΔHrxn is negative.
The enthalpy of reaction for the combustion of CH4, the main component in natural gas:
This reaction is exothermic and therefore has a negative enthalpy of reaction.
The magnitude of ΔHrxn tells us that kJ of heat are emitted when 1 mol CH4 reacts with 2 mol O2.

43. Sign of ΔHrxn
When thermal energy flows into the reaction and out of the surroundings (as in an endothermic reaction), then ΔHrxn is positive.
The enthalpy of reaction for the reaction between nitrogen and oxygen gas to form nitrogen monoxide:
This reaction is endothermic and therefore has a positive enthalpy of reaction.
The magnitude of ΔHrxn tells us that kJ of heat are absorbed from the surroundings when 1 mol N2 reacts with 1 mol O2.

44. Stoichiometry of ΔHrxn
The amount of heat emitted or absorbed when a chemical reaction occurs depends on the amounts of reactants that actually react.
We usually specify ΔHrxn in combination with the balanced chemical equation for the reaction.
The magnitude of ΔHrxn is for the stoichiometric amounts of reactants and products for the reaction as written.

45. Stoichiometry of ΔHrxn
For example, the balanced equation and ΔHrxn for the combustion of propane (the fuel used in LP gas) is:
When 1 mole of C3H8 reacts with 5 moles of O2 to form 3 moles of CO2 and 4 moles of H2O, 2044 kJ of heat are emitted.
These ratios can be used to construct conversion factors between amounts of reactants or products and the quantity of heat exchanged.

46. Stoichiometry of ΔHrxn
To find out how much heat is emitted upon the combustion of a certain mass in grams of propane C3H8, we can use the following solution map:

47. Example 8.7 Stoichiometry Involving ΔHrxn
An LP gas tank in a home barbecue contains 11.8 x 103 g of propane (C3H8).
Calculate the heat (in kJ) associated with the complete combustion of all of the propane in the tank.

48. 1 mol C3H8 : –2044 kJ (from balanced equation)
Example: Complete combustion of x 103 g of propane (C3H8) SOLUTION MAP:
RELATIONSHIPS USED:
1 mol C3H8 : –2044 kJ (from balanced equation)
Molar mass C3H8 = g/mol

49. Example: Complete combustion of 11.8 x 103 g of propane (C3H8)
SOLUTION:
Often in the homework, the absolute value of Q, |Q|, is requested and words are used to convey the sign of the heat absorbed or given off in the reaction.

50. Everyday Chemistry Bunsen Burners
Most Bunsen burners have a mechanism to adjust the amount of air (and therefore of oxygen) that is mixed with the methane.
If you light the burner with the air completely closed off, you get a yellow, smoky flame that is not very hot.
As you increase the amount of air going into the burner, the flame becomes bluer, less smoky, and hotter.
When you reach the optimum adjustment, the flame has a sharp, inner blue triangle, gives off no smoke, and is hot enough to melt glass easily.
Continuing to increase the air beyond this point causes the flame to become cooler again and may actually extinguish it.

51. A Bunsen burner at various stages of air intake adjustment.