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Chapter 12 Stoichiometry Quantity. 12.1 Stoichiometry Chemical reactions represent the heart of chemistry: they describe the endless ways that substances.

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Presentation on theme: "Chapter 12 Stoichiometry Quantity. 12.1 Stoichiometry Chemical reactions represent the heart of chemistry: they describe the endless ways that substances."— Presentation transcript:

1 Chapter 12 Stoichiometry Quantity

2 12.1 Stoichiometry Chemical reactions represent the heart of chemistry: they describe the endless ways that substances can combine with each other to form new substances. In essence, they describe what chemistry really is -- the study of matter and its transformations through Chemical Reactions.

3 Ch 12 Stoichiometry is the area of chemistry which deals with quantitative relationships in chemical reactions. It is the area which allows chemists to know how much starting material is needed to produce a million pounds of sulfuric acid or how many molecules are in a microgram of a particular hormone.

4 When I worked in a semi trailer factory, guess who was highest paid? Not the factory worker (Minimum wage) Not the welders (4 times minimum wage) Not the foreman (5 times minimum wage) It was the warehouse manager (6 times minimum wage) Why??? He had to know how to do Stoichiometry to produce the correct amount of chemicals needed for production.

5 Stoichiometry Predicting the quantities of products when given the quantities of reagents.

6 Stoichiometry: Quantities are usually given in grams, kilograms, or pounds. Are chemical equations in grams?

7 Stoichiometry Predicting the quantities of products when given the quantities of reagents. Quantities are usually given in grams, kilograms, or pounds. Are chemical equations in grams? No, in representative particles or moles. Usually we have to: 1. Convert from grams to moles. 2. Work equations with moles. 3. Convert moles back to grams.

8 What is the mass of 2 moles of F 2 ? How much is in 2 moles of F 2 ? (written as 2 F 2 ) Moles = 2 moles of F 2 Molecules = 2 molecules of F 2 (or 2 x 6.02 x molecules) Atoms = 4 atoms F (or 4 x 6.02 x atoms) Grams = 76g (F has atomic mass of 19) (total grams in 2 moles of F 2 ) Liters = 44.8L (C.F. = 22.4 L per Mole at STP)

9 Stoichiometry (“stoy-key-AHM-uh-tree”) The relationship between the quantities of reactants and products in a chemical reaction (Eqn next page.)

10 P 4(s) + 6 Cl 2(g)  4 PCl 3(s) From the balanced equation the following information can be obtained.

11 U try it! 2Na + Cl 2  2NaCl Moles? Grams? Liters?

12 U try it! 2Na + Cl 2  2NaCl Moles? Grams? Liters? 22.4

13 Mole ratios The ratio of moles given by the balanced equation. N 2 + 3F 2  2NF 3 For every 1 mole of N 2 there are 2 moles of NF 3 formed (assuming that there is at least 3 F 2 ) Mole Ratios: N 2 :NF 3 is 1 to 2 F 2 :NF 3 is 3 to 2

14 Mole to Mole calculations How many moles of ozone O 3 will be formed from 6 moles of oxygen O 2 ?

15 O 2  O 3 Step 1 balance the equation.

16 How many moles of ozone O 3 will be formed from 6 moles of oxygen O 2 ? Step 2 Identify what the question is asking… Goal? –List Known’s and Unknown’s

17 How many moles of ozone O 3 will be formed from 6 moles of oxygen O 2 ? Step 3 Use dimensional analysis to answer solve the problem

18 How many moles of ozone O 3 will be formed from 6 moles of oxygen O 2 ? Check your answer to see if it is logical.

19 Can you make the six mole ratios? 4Al + 3O 2  2Al 2 O 3

20 How many moles of aluminum and oxygen are needed to form 3.21 moles of aluminum oxide? 4Al + 3O 2  2Al 2 O 3

21 How many moles of chlorine are needed to make 15 moles of NaCl?

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24 C + O 2  CO 2 1 C atom reacts with 1 O 2 molecule to form 1 CO 2 molecule 12 C atoms reacts with 12 O 2 molecules to form 12 CO 2 molecules 1 mol C atoms reacts with 1 mol O 2 molecules to form 1 mol CO 2 molecules What do you need to make 12 mol CO 2 ?

25 12.2 Chemical calculations Start here!

26 12.2 Chemical calculations

27 1 mole = 6.02 x representative particles = molar mass (grams) = 22.4L (gases only)

28 Mole ratios 2 moles of O 3 form 3 moles of O 2 2 O 3  3 O 2

29 (Gram, molecules, particles, atoms) –Remember ALWAYS convert to moles! –Then use mole ratio.

30 How many grams of water can be formed from 15g of hydrogen? (assume there is unlimited oxygen) 2H 2 + O 2  2H 2 O –g  mol hydrogen Then, mole ratio to find water then convert moles of water  grams

31 2H 2 + O 2  2H 2 O

32 How many H 2 molecules are needed to make 50g of water? (assume there is unlimited oxygen) 2H 2 + O 2  2H 2 O

33 50g of water 2H 2 + O 2  2H 2 O –g H 2 O  moles H 2 O Moles H 2 O  moles H 2 –Moles H 2  molecules of H 2

34 Greenhouse gases North America, consists of the United States and Canada. North America is the highest fossil-fuel, CO 2 emitting region of the world with 1.73 billion tons of carbon in This 2002 total is an all- time high for North America and represents a 1.4% increase from Because ~92% of current fossil-fuel CO 2 emissions from the region are from the United States, the North America data closely resembles that for the United States.

35 How many L of CO 2(g) will be formed from 16 gallons of gasoline (64,000g)? (assume gas is 100% octane) 2C 8 H O 2  16CO H 2 O

36 197,614g = 2.0 x ,000L of CO 2 made for each 16 gal of gas burned.

37 12.5 g Li  g product O 2 Li + O 2  Li 2 O S Li + S  Li 2 S

38 12.5 g Li  Predict g product O 2 Li + O 2  Li 2 O S Li + S  Li 2 S

39 12.5 g Li  Predict g product Step 1, Balance O 2 Li + O 2  Li 2 O 4Li + 2O 2  2Li 2 O S Li + S  Li 2 S 2Li + S  Li 2 S

40 Use last problem to intro 12.3

41 12.5 g Li  g product What is your a.)Theoretical Yield b.) Actual Yield c.) Percent Yield Li + O 2  Li 2 O Previous Ans: You calculate 26.9 g Li 2 O You measure the product and there is 25 grams of Li 2 O Li + S  Li 2 S Previous Ans: You calculate 41.4 g Li 2 S You measure the product and there is 37.0 grams of Li 2 S

42 12.5 g Li  g product Li + O 2  Li 2 O a.)Theoretical Yield 27.0 g Li 2 O b.) Actual Yield 25 g Li 2 O c.) Percent Yield 25/26.9 x100=92.9% Li + S  Li 2 S a.)Theoretical Yield 41.4 g Li 2 S b.) Actual Yield 37.0 g Li 2 S c.) Percent Yield 37.0/41.4 x100=89.4%

43 Actual Yield = MEASURED amount (recorded by experiment) Theoretical Yield = CALCULATED amount (What balance equation says we should have produced) Percent Yield is ratio of GRAMS!

44 12.3 limiting reagents limiting reagent – reactant that will be used up first and cause the reaction to stop producing products

45 Identify the limiting reagent Example Problem (solved on next slide) 2Na + Cl 2  2NaCl –Givens are starting amounts of each reagent –Which is the limiting reactant? STEPS to find limiting reagent (Balance equation.) 1.) Calculate amount of product produced by each reagent. 2.) Limiting reagent is the one that produces the least amount of product. 3.) Excess reagent is the leftover amounts of the other reagent

46 Identify the limiting reagent 2Na + Cl 2  2NaCl –1 mol Na reacts with 1 mol Cl 2? –Which is the limiting reactant?

47 Identify the excess reagent 2Na + Cl 2  2NaCl –1 mol Na reacts with 1 mol Cl 2? –How much of excess reagent is there?

48 What is the actual equation? 1 Na + 1 Cl 2  1 NaCl + ½ Cl 2 Limiting Excess Reagent

49 Identify the limiting reagent N 2 + O 2  NO 3 –1.7 mol N 2 or 1.3 mol O 2

50 Identify the excess reagent. N 2 + O 2  NO 3 –1.7 mol N 2 or 1.3 mol O 2

51 58g of Na are reacted with 110g of S how much product will be formed? 2Na + S  Na 2 S Discuss steps

52 What is the percent yield if 170g of sodium sulfide was produced? (see previous problem) Because of human error, measurement error, contaminated reactants, experimental error, ect… there will usually be less than 100% yield, meaning that less products will be produced then you calculated for.

53 15.5g of H 2 is reacted with 155g of O 2. What is the percent yield if 136g of H 2 O has been collected? 2H 2 + O 2  2H 2 O

54 ANSWER: 15.5g of H 2 is reacted with 155g of O 2. What is the percent yield if 136g of H 2 O has been collected? 2H 2 + O 2  2H 2 O 97.1%

55 SUMMARY Write the balanced equation. Convert variables to moles. Find the moles of each reactant needed based on the amount of the other reactant given (use Na to find S). Identify the limiting reagent. Solve for the amounts of reactants needed and products formed.


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