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Chapter #8 Chemical Quantities. Chapter Outline 8.2 Mole relationships defined in balanced equations 8.3 Mole conversions- Stoichiometry 8.4 Mass Conversions.

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Presentation on theme: "Chapter #8 Chemical Quantities. Chapter Outline 8.2 Mole relationships defined in balanced equations 8.3 Mole conversions- Stoichiometry 8.4 Mass Conversions."— Presentation transcript:

1 Chapter #8 Chemical Quantities

2 Chapter Outline 8.2 Mole relationships defined in balanced equations 8.3 Mole conversions- Stoichiometry 8.4 Mass Conversions Stoichiometry 8.5 Excess and Limiting Reactants 8.5 Theoretical Yield 8.6 Theoretical Yield From Initial Reactant Masses 8.7 Thermochemical Equations

3 STOICHIOMETRY Stoichiometry is the use of balanced chemical equations in the conversion process.

4 8.4 STOICHIOMETRY Stoichiometry is the use of balanced chemical equations in the conversion process. Examples Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required. 2 H 2 + O 2 2H 2 O

5 8.4 STOICHIOMETRY Stoichiometry is the use of balanced chemical equations in the conversion process. Examples Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required. 2 H 2 + O 2 2H 2 O 6.33 g H 2

6 8.4 STOICHIOMETRY Stoichiometry is the use of balanced chemical equations in the conversion process. Examples Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required. 2 H 2 + O 2 2H 2 O 6.33 g H g H 2 mole H 2

7 8.4 STOICHIOMETRY Stoichiometry is the use of balanced chemical equations in the conversion process. Examples Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required. 2 H 2 + O 2 2H 2 O 6.33 g H g H 2 Mole H 2

8 8.4 STOICHIOMETRY Stoichiometry is the use of balanced chemical equations in the conversion process. Examples Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required. 2 H 2 + O 2 2H 2 O 6.33 g H g H 2 O Mole H 2 2 mole H 2 O 2 Mole H 2

9 8.4 STOICHIOMETRY Stoichiometry is the use of balanced chemical equations in the conversion process. Examples Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required. 2 H 2 + O 2 2 H 2 O 6.33 g H g H 2 Mole H 2 2 mole H 2 O 2 Mole H 2 Mole H 2 O g H 2 O

10 8.4 STOICHIOMETRY Stoichiometry is the use of balanced chemical equations in the conversion process. Examples Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required. 2 H 2 + O 2 2 H 2 O 6.33 g H g H 2 Mole H 2 2 mole H 2 O 2 Mole H 2 Mole H 2 O g H 2 O = 28.3 g H 2 O

11 8.5 Excess and Limiting Reactants Reactants are substances that can be changed into something else. For example, nails and boards are reactants for carpenters, while thread and fabric are reactants for the seamstress. And for a chemist hydrogen and oxygen are reactants for making water.

12 8.5 Building Houses Ok, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make?

13 8.5 Building Houses Ok, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make? Yes, only one house!

14 8.5 Building Houses Ok, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make? What reactant is in excess? And how many more houses could we use if we had enough boards?

15 8.5 Building Houses Ok, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make? What reactant is in excess? And how many more houses could we use if we have enough boards?

16 8.5 Building Houses Ok, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make? What reactant is in excess? And how many more houses could we use if we have enough boards? Yes, nails are in excess!

17 8.5 Building Houses Ok, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make? What reactant is in excess? And how many more houses could we use if we have enough boards? Yes, nails are in excess! Nine more houses if we have an adequate amount of boards.

18 8.5 Excess and Limiting Example If we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess?

19 8.5 Excess and Limiting Example If we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess? Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen.

20 8.5 Excess and Limiting Example If we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess? Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen. 2 H 2 + O 2 2 H 2 O 10.0 g O 2

21 8.5 Excess and Limiting Example If we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess? Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen. 2 H 2 + O 2 2 H 2 O 10.0 g O g O 2 mole O 2

22 8.5 Excess and Limiting Example If we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess? Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen. 2 H 2 + O 2 2 H 2 O 10.0 g O g O 2 mole O 2 2 mole H 2

23 8.5 Excess and Limiting Example If we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess? Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen. 2 H 2 + O 2 2 H 2 O 10.0 g O g O 2 mole O 2 2 mole H 2 mole H g H 2

24 8.5 Excess and Limiting Example If we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess? Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen. 2 H 2 + O 2 2 H 2 O 10.0 g O g O 2 mole O 2 2 mole H 2 mole H g H 2 = 1.26 g H 2

25 8.5 Excess and Limiting Example Only 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water.

26 8.5 Excess and Limiting Example Only 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water g O 2 mole O g O 2

27 8.5 Excess and Limiting Example Only 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water g O 2 mole O g O 2 mole O 2 2 mole H 2 O

28 8.5 Excess and Limiting Example Only 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water g O 2 mole O g O 2 mole O 2 2 mole H 2 O 18.0 g H 2 O mole H 2 O

29 8.5 Excess and Limiting Example Only 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water g O 2 mole O g O 2 mole O 2 2 mole H 2 O 18.0 g H 2 O mole H 2 O = 11.3 g H 2 O

30 8.5 Percentage Yield The percent yield is a comparison of the laboratory answer to the correct answer which is determined by the conversion process. Suppose a student combined 10.0 g of oxygen and 10.0 g of hydrogen in the lab and recovered 8.66 g of water. What would be the percent yield?

31 8.5 Percentage Yield The percent yield is a comparison of the laboratory answer to the correct answer which is determined by the conversion process. Suppose a student combined 10.0 g of oxygen and 10.0 g of hydrogen in the lab and recovered 8.66 g of water. What would be the percent yield? percent yield = Yield (the lab amount) Theoretical Yield (by conversions) X 100 percent yield = X 100 = 76.6%

32 8.7 Thermochemistry Exothermic Reaction Endothermic Reaction - ΔH + ΔH E act

33 8.7 Thermochemistry When a chemical or physical change takes place energy is either lost of gained. A Thermochemical equation describes this change. Equations gaining energy are called endothermic and equations losing energy are called exothermic.

34 8.7 Thermochemical Equations When a chemical or physical change takes place energy is either lost of gained. A Thermochemical equation describes this change. Equations gaining energy are called endothermic and equations losing energy are called exothermic. Examples: C 3 H 6 O (l ) + 4O 2 (g) 3CO 2 (g) + 3 H 2 O (g) ΔH = kj Exothermic H 2 O (l) H 2 O (g) ΔH = kj Endothermic

35 8.7 Thermochemical Equations How many kj of heat are released when 709 g of C 3 H 6 O are burned?

36 8.7 Thermochemical Equations How many kj of heat are released when 709 g of C 3 H 6 O are burned? 709 g C 3 H 6 O C 3 H 6 O (l ) + 4O 2 (g) 3CO 2 (g) + 3 H 2 O (g) ΔH = kj

37 8.7 Thermochemical Equations How many kj of heat are released when 709 g of C 3 H 6 O are burned? 709 g C 3 H 6 O C 3 H 6 O (l ) + 4O 2 (g) 3CO 2 (g) + 3 H 2 O (g) ΔH = kj 58.1 g C 3 H 6 O mole C 3 H 6 O

38 8.7 Thermochemical Equations How many kj of heat are released when 709 g of C 3 H 6 O are burned? 709 g C 3 H 6 O C 3 H 6 O (l ) + 4O 2 (g) 3CO 2 (g) + 3 H 2 O (g) ΔH = kj 58.1 g C 3 H 6 O mole C 3 H 6 O

39 8.7 Thermochemical Equations How many kj of heat are released when 709 g of C 3 H 6 O are burned? 709 g C 3 H 6 O C 3 H 6 O (l ) + 4O 2 (g) 3CO 2 (g) + 3 H 2 O (g) ΔH = kj 58.1 g C 3 H 6 O mole C 3 H 6 O 1790 kj = kj

40 The End


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