1. Quantities in Chemical Reactions Chemical Principles Stoichiometry Stoichiometry Limiting Reactant, Theoretical Yield, and Percent Yield Limiting Reactant, Theoretical Yield, and Percent Yield

2. STOICHIOMETRY Definition Time: STOICHIOMETRY the mathematical relationship between any two chemicals in a balanced chemical equation the mathematical relationship between any two chemicals in a balanced chemical equation Earlier in this course we learned how to balance equations and to convert between grams and moles. Earlier in this course we learned how to balance equations and to convert between grams and moles. Now we are going to combine the two processes. Now we are going to combine the two processes.

3. Quantities in Chemical Reactions Making Pancakes: Relationships Similar to Recipe 5 Pancakes 1 cup of Flour 2 Eggs ½ tsp Baking Powder

4. Quantities in Chemical Reactions Making Pancakes: Relationships Numerical Relationship Numerical Relationship 2 Eggs  5 Pancakes 2 Eggs  5 Pancakes Two eggs are required to produce 5 pancakes. Two eggs are required to produce 5 pancakes. Numerical Relationship for Ingredients 1 cup Flour  5 Pancakes ½ tsp Baking Powder  5 Pancakes 1 cup Flour  ½ tsp Baking Powder

5. Quantities in Chemical Reactions Making Molecules: Mole to Mole Balanced Chemical Equation Balanced Chemical Equation Recipe Recipe 3 H 2 (g) + N 2 (g)  2 NH 3 (g) 3 H 2 (g) + N 2 (g)  2 NH 3 (g) Possible Mole Ratios? 3 moles H 2 (g)  1 mole N 2 (g)  2 moles NH 3 (g)

6. Quantities in Chemical Reactions Making Molecules: Mole to Mole 3 H 2 (g) + N 2 (g)  2 NH 3 (g) 3 H 2 (g) + N 2 (g)  2 NH 3 (g) Mole Ratios? 3 moles H 2 (g)  1 mole N 2 (g) which is 3:1 or 3/1 3 moles H 2 (g)  2 moles NH 3 (g) which is 3:2 or 3/2 1 mole N 2 (g)  2 moles NH 3 (g) which is 1:2 or 1/2 So… 6 moles H 2 (g)  ? mole N 2 (g) 5 moles H 2 (g)  ? mole N 2 (g) 3 mole N 2 (g)  ? moles NH 3 (g)

7. Quantities in Chemical Reactions Making Molecules: Mole to Mole 3 H 2 (g) + N 2 (g)  2 NH 3 (g) 3 H 2 (g) + N 2 (g)  2 NH 3 (g) Mole Ratios? 2 moles N 2 (g)  ? mole H 2 (g) 5 moles NH 3 (g)  ? mole N 2 (g) 3 mole NH 3 (g)  ? moles N 2 (g)

8. STOICHIOMETRY We have already done most of the TYPES of problems that we will see in this section. We are going to put several short problems together to make longer problems. Performing stoichiometry calculations going from is a 3 step process: 1)convert what you are given into moles, 1)convert what you are given into moles, 2)use the mole to mole ratio from the BALANCED EQUATION to relate the two chemicals 2)use the mole to mole ratio from the BALANCED EQUATION to relate the two chemicals 3)convert the answer to the units requested in the problem (usually grams) 3)convert the answer to the units requested in the problem (usually grams)

9. STOICHIOMETRY The easiest way to learn stoichiometry is to do problems: How many grams of NaCl would be produced if 100. grams of NaOH react with excess HCl according to the following equation: NaOH + HCl  NaCl + H 2 O How many grams of NaCl would be produced if 100. grams of NaOH react with excess HCl according to the following equation: NaOH + HCl  NaCl + H 2 O

10. STOICHIOMETRY Before you even begin the problem, you should make sure the equation is balanced. Before you even begin the problem, you should make sure the equation is balanced. NaOH + HCl  NaCl + H 2 O NaOH + HCl  NaCl + H 2 O Yes, the equation is balanced. Yes, the equation is balanced.

11. STOICHIOMETRY An overall look at the steps involved: NaOH + HCl  NaCl + H 2 O NaOH + HCl  NaCl + H 2 O Moles of NaOH Moles of NaCl Use MW Of NaOH Use MW Of NaCl Use Mole to Mole Ratio

12. How many grams of NaCl would be produced if 100. grams of NaOH react with excess HCl according to the following equation: NaOH + HCl  NaCl + H 2 O Step 1: Convert what you are given to moles:

13. How many grams of NaCl would be produced if 100 grams of NaOH react with excess HCl according to the following equation: NaOH + HCl  NaCl + H 2 O Step 2) Use mole to mole ratio: We will use the ratio between NaCl (what the problem asks us to find) and the NaOH (what the problem gives us to start with). Be sure to write all units and types of chemicals. We will use the ratio between NaCl (what the problem asks us to find) and the NaOH (what the problem gives us to start with). Be sure to write all units and types of chemicals.

14. How many grams of NaCl would be produced if 100 grams of NaOH react with excess HCl according to the following equation: NaOH + HCl  NaCl + H 2 O Step 3) Convert to the units asked for in the problem: The problem asks for number of grams of NaCl. So we need to convert the moles of NaCl to grams. The problem asks for number of grams of NaCl. So we need to convert the moles of NaCl to grams. 2.5 moles NaCl x 58.5 g/mole = 146g NaCl 2.5 moles NaCl x 58.5 g/mole = 146g NaCl

15. STOICHIOMETRY How about another problem? How about another problem? Using the equation given below, determine how many grams of Na 3 PO 4 would be needed to react with 5.5 grams of Ba(NO 3 ) 2 Using the equation given below, determine how many grams of Na 3 PO 4 would be needed to react with 5.5 grams of Ba(NO 3 ) 2 Ba(NO 3 ) 2 + Na 3 PO 4  Ba 3 (PO 4 ) 2 + NaNO 3 Ba(NO 3 ) 2 + Na 3 PO 4  Ba 3 (PO 4 ) 2 + NaNO 3

16. STOICHIOMETRY Although it is not one of the OFFICIAL steps, we should always make sure the equation is balanced BEFORE we begin the problem. Remember, stoichiometry is the relationship between two chemicals in a BALANCED equation. 3Ba(NO 3 ) 2 + 2Na 3 PO 4  Ba 3 (PO 4 ) 2 + 6 NaNO 3 3Ba(NO 3 ) 2 + 2Na 3 PO 4  Ba 3 (PO 4 ) 2 + 6 NaNO 3 Determine how many grams of Na 3 PO 4 would be needed to react with 5.50 grams of Ba(NO 3 ) 2. Determine how many grams of Na 3 PO 4 would be needed to react with 5.50 grams of Ba(NO 3 ) 2.

17. 3Ba(NO 3 ) 2 + 2Na 3 PO 4  1Ba 3 (PO 4 ) 2 + 6 NaNO 3 Step 1) Convert 5.50 grams Ba(NO 3 ) 2 to moles: Step 2) Use mole to mole ratio from balanced equation: Step 3) Convert to the proper units: So, the answer is “2.3 grams of sodium phosphate are required to react with 5.5 grams of barium nitrate.”

18. STOICHIOMETRY Sometimes you may not have to do all three steps--when you READ the problem it will give you clues. Sometimes you may not have to do all three steps--when you READ the problem it will give you clues. How many moles of Hydrogen gas will be produced when 2 moles of sodium react according to the following balanced equation? How many moles of Hydrogen gas will be produced when 2 moles of sodium react according to the following balanced equation?

19. 2Na + 2H 2 O  H 2 + 2NaOH The problem gives us some clues such as telling us that the equation is balanced. Step 1) Convert to moles: We are given 2 moles of sodium so this step is not necessary. Step 2) Use mole ratio from balanced equation: Using the balanced equation… 2 moles X 1 mole H 2 = 1 mole H 2 2 moles X 1 mole H 2 = 1 mole H 2 2 mole Na Step 3) Convert to units requested in problem: The problem asked for moles of hydrogen gas--we already have that. The answer is 1 mole of hydrogen gas.

20. STOICHIOMETRY There are a few key points to remember about Stoichiometry. There are a few key points to remember about Stoichiometry. Stoichiometry can only look at TWO chemicals at one time Stoichiometry can only look at TWO chemicals at one time It does not matter if the two chemicals are  both products,  both reactants,  or one of each. Stoichiometry problems can get harder than the examples.

21. LIMITING REACTANT The Limiting Reactant is defined as the reactant that runs out first in a chemical reaction. The Limiting Reactant is defined as the reactant that runs out first in a chemical reaction. 1 cup of Flour+2 Eggs +½ tsp Baking Powder  5 pancakes 3 cups of Flour, 10 eggs, and 4 tsp baking powder 3 cups of Flour, 10 eggs, and 4 tsp baking powder 3 cups of flour  15 pancakes 3 cups of flour  15 pancakes 10 eggs  25 pancakes 10 eggs  25 pancakes 4 tsp baking powder  40 pancakes 4 tsp baking powder  40 pancakes

22. LIMITING REACTANT 1 cup of Flour + 2 Eggs + ½ tsp Baking Powder  5 pancakes 3 cups of flour, 10 eggs, and 4 tsp baking powder 3 cups of flour, 10 eggs, and 4 tsp baking powder 3 cups of flour  15 pancakes 3 cups of flour  15 pancakes 10 eggs  25 pancakes 10 eggs  25 pancakes 4 tsp baking powder  40 pancakes 4 tsp baking powder  40 pancakes

23. LIMITING REACTANT 1 cup of Flour + 2 Eggs + ½ tsp Baking Powder  5 pancakes 3 cups of flour  15 pancakes 3 cups of flour  15 pancakes Limiting Reactant (Reagent) Limiting Reactant (Reagent) Produces the Least Amount of Product Produces the Least Amount of Product

24. Limiting Reactant Example: If we begin with 1.8 moles of titanium and 3.2 moles of chlorine, what is the limiting reactant in moles of TiCl 4 ? Ti(s) + 2 Cl 2 (g)  TiCl 4 (s) Ti(s) + 2 Cl 2 (g)  TiCl 4 (s) Given: 1.8 moles Ti and 3.2 moles Cl 2 Find: Limiting reactant Conversions: 1 mole Ti  1 mole TiCl 4 2 mole Cl 2  1 mole TiCl 4

25. Limiting Reactant If we begin with 1.8 moles of titanium and 3.2 moles of chlorine, what is the limiting reactant in moles of TiCl 4 ? (We are given information about TWO reactants and asked to find ONE of the products.) Ti(s) + 2 Cl 2 (g)  TiCl 4 (s) Ti(s) + 2 Cl 2 (g)  TiCl 4 (s) 1.8 moles X 1 mole TiCl 4 = 1.8 moles TiCl 4 1 mole Ti 1 mole Ti 3.2 moles X 1 mole TiCl 4 = 1.6 moles TiCl 4 2 mole Cl 2 2 mole Cl 2 Least Amount of Product Limiting Reactant

26. LIMITING REACTANT – Using grams Example: 25 grams of HCl are mixed with 25 grams of NaOH. How many grams of NaCl can be made based on the following balanced chemical equation: 25 grams of HCl are mixed with 25 grams of NaOH. How many grams of NaCl can be made based on the following balanced chemical equation: HCl + NaOH  NaCl + H 2 O HCl + NaOH  NaCl + H 2 O

27. LIMITING REACTANT – Using grams Limiting reactant problems are not any harder than normal stoichiometry problems, they are just twice as long. We need to do TWO stoichiometry problems. Limiting reactant problems are not any harder than normal stoichiometry problems, they are just twice as long. We need to do TWO stoichiometry problems. The first problem will relate the NaOH to the NaCl. The first problem will relate the NaOH to the NaCl. The second problem will relate the HCl to the NaCl The second problem will relate the HCl to the NaCl

28. LIMITING REACTANT – Using grams After we have finished the two problems, we will obviously have two answers. After we have finished the two problems, we will obviously have two answers. The SMALLER of the two answers will be the limiting amount of NaCl that is produced and the reactant that produces the smaller number will be the limiting reactant (it will run out first). The SMALLER of the two answers will be the limiting amount of NaCl that is produced and the reactant that produces the smaller number will be the limiting reactant (it will run out first).

29. HCl + NaOH NaCl + H 2 O PROBLEM #1 Step 1--convert 25 grams of NaOH to moles… 25 grams x 1 mole/40 grams = 0.63 moles 25 grams x 1 mole/40 grams = 0.63 moles Step 2--use mole to mole ratio…. 0.63 moles NaOH x 1 mole NaCl = =0.63 moles of NaCl 0.63 moles NaOH x 1 mole NaCl = =0.63 moles of NaCl 1 mole NaOH 1 mole NaOH Step 3--convert to grams… 0.63 moles NaCl x 58.5 g/mole = 36.9 grams of NaCl 0.63 moles NaCl x 58.5 g/mole = 36.9 grams of NaCl This is our first answer. This is our first answer. Now we can go to the second problem Now we can go to the second problem Problem #1 Problem #2

30. HCl + NaOH NaCl + H 2 O PROBLEM #2 Step 1--convert 25 grams HCl to moles… 25 grams x 1 mole = 0.68 moles of HCl 25 grams x 1 mole = 0.68 moles of HCl 36.5 grams 36.5 grams Step 2--use mole ratio… 0.68 moles HCl x 1 mole NaCl = 0.68 moles NaCl 0.68 moles HCl x 1 mole NaCl = 0.68 moles NaCl 1 mole HCl 1 mole HCl Step 3--convert to grams… 0.68 moles x 58.5g/mole = 39.8 g NaCl 0.68 moles x 58.5g/mole = 39.8 g NaCl Now we have our second answer. Now we have our second answer. Problem #1 Problem #2

31. HCl + NaOH  NaCl + H 2 O The First Problem, looking at the relationship between NaOH and NaCl, gave us an answer of 36.9 grams of NaCl The First Problem, looking at the relationship between NaOH and NaCl, gave us an answer of 36.9 grams of NaCl The Second Problem, looking at the relationship between HCl and NaCl gave us an answer of 39.8 grams of NaCl n The first number is smaller.

32. HCl + NaOH  NaCl + H 2 O The NaOH is the limiting reactant because it produced the smaller answer. The NaOH is the limiting reactant because it produced the smaller answer. The amount of NaCl that was produced in the reaction was 36.9 grams. The amount of NaCl that was produced in the reaction was 36.9 grams. But wait, there is more. But wait, there is more.

33. HCl + NaOH  NaCl + H 2 O The problem also may ask-- “How many grams of the NON-Limiting reactant will be left over at the end of the problem.” “How many grams of the NON-Limiting reactant will be left over at the end of the problem.” HCl is the NON-limiting reactant. HCl is the NON-limiting reactant. Now we are back to a basic stoichiometry problem…. Now we are back to a basic stoichiometry problem….

34. HCl + NaOH  NaCl + H 2 O We can determine how much HCl was CONSUMED reacting with the 25 grams of NaOH. We can determine how much HCl was CONSUMED reacting with the 25 grams of NaOH. Once we know how much was consumed, we can subtract this from the 25 grams we started with to determine how much was left over. Once we know how much was consumed, we can subtract this from the 25 grams we started with to determine how much was left over.

35. HCl + NaOH  NaCl + H 2 O This is a simple three-part stoichiometry problem. Step 1--convert to grams 25grams NaOH x 1 mole/40grams = 0.63mole of NaOH Step 2--mole to mole ratio 0.63moles NaOH x 1mole HCl/1mole NaOH = 0.63 moles of HCl Step 3--convert to grams 0.63 moles X 36.5 g/mole = 23 grams of HCl This HCl was CONSUMED

36. HCl + NaOH  NaCl + H 2 O Since we started with 25 grams of HCL and we used 23 grams of HCl, there will be 2 grams of HCl leftover. Since we started with 25 grams of HCL and we used 23 grams of HCl, there will be 2 grams of HCl leftover. The next part of the question may ask: The next part of the question may ask: How many grams of the limiting reactant are left over? How many grams of the limiting reactant are left over?

37. HCl + NaOH  NaCl + H 2 O This is an easy stoichiometry problem. This is an easy stoichiometry problem. The answer is 0 grams. The answer is 0 grams. The limiting reactant is the one that runs out first therefore it can’t have leftovers. The limiting reactant is the one that runs out first therefore it can’t have leftovers. Be sure to read the problem and answer the problem the teacher asks and not the one that you think the teacher is going to ask.

38. HCl + NaOH  NaCl + H 2 O The last question is: The last question is: How many grams of water will be produced? How many grams of water will be produced? There are four ways to do this problem. There are four ways to do this problem. 1)relate HCl to H 2 O 2)relate NaOH to H 2 O 3)relate NaCl to H 2 O 4)use the Law of Conservation of Mass.

39. HCl + NaOH  NaCl + H 2 O We can do this last problem four different ways and see if we get the same answer… We can do this last problem four different ways and see if we get the same answer…

40. HCl + NaOH  NaCl + H 2 O What do we know from previous problems? What do we know from previous problems? 25 grams of NaOH is the limiting reactant 25 grams of NaOH is the limiting reactant 23 grams of HCl was also used 23 grams of HCl was also used 36.9 grams of NaCl was produced. 36.9 grams of NaCl was produced. We will do the problems with a minimum of discussion to save time and space.

41. HCl + NaOH  NaCl + H 2 O Calculations from 23grams of HCl: Calculations from 23grams of HCl: n Calculations from 25grams of NaOH: Calculations from 36.9 grams of NaCl:

42. HCl + NaOH  NaCl + H 2 O Law of Conservation of Mass We used 23 grams of HCl and 25 grams of NaOH. Therefore, the total mass of the reactants that was used was 48 grams. We used 23 grams of HCl and 25 grams of NaOH. Therefore, the total mass of the reactants that was used was 48 grams. The Law of Conservation of Mass tells us that what we start with is what we end with. The Law of Conservation of Mass tells us that what we start with is what we end with. Therefore, we must end with 48 grams of products. We ended with 36.9 grams of NaCl, so we must have had 11.1 grams of H 2 O based on Conservation of Mass. Therefore, we must end with 48 grams of products. We ended with 36.9 grams of NaCl, so we must have had 11.1 grams of H 2 O based on Conservation of Mass. (Remember--the other two grams of HCl that we DID NOT use does not count either way.) (Remember--the other two grams of HCl that we DID NOT use does not count either way.)

43. HCl + NaOH NaCl + H 2 O We now have four answers: We now have four answers: Starting with HCl we ended up with 11.3 grams of H 2 O Starting with HCl we ended up with 11.3 grams of H 2 O Starting with NaOH we ended up with 11.3 grams Starting with NaOH we ended up with 11.3 grams Starting with NaCl we ended up with 11.3 grams, Using the Law of Conservation of Mass, we ended up with 11.1 grams. They are different-- but not much!

44. HCl + NaOH NaCl + H 2 O The reason for the difference is the rounding that took place during the problem. The reason for the difference is the rounding that took place during the problem. If you remember sig figs, we should only have 2 sig figs anyway based on the 25 grams of starting material. If you remember sig figs, we should only have 2 sig figs anyway based on the 25 grams of starting material. Using correct sig fig’s, all methods give the same answer = 11 grams of water. Using correct sig fig’s, all methods give the same answer = 11 grams of water.

45. YIELD CALCULATIONS Let’s go to another topic... THEORETICAL YIELD is the mathematical answer produced based solely on calculations. ACTUAL YIELD is the yield produced in the lab when someone actually does the experiment.

46. More Pancakes: Limiting Reactant, Theoretical Yield, and Percent Yield 1 cup of Flour + 2 Eggs + ½ tsp Baking Powder  5 pancakes Theoretical Yield Theoretical Yield 15 pancakes  Most that can be produced 15 pancakes  Most that can be produced Actual Yield Actual Yield What was actually produced  11 pancakes What was actually produced  11 pancakes Burned 3 pancakes Burned 3 pancakes Dropped 1 pancake Dropped 1 pancake Percent Yield Percent Yield Percent of the theoretical yield actually attained Percent of the theoretical yield actually attained Percent Yield Percent Yield Percent of the theoretical yield actually attained Percent of the theoretical yield actually attained = 11 pancakes X 100 = 73% 15 pancakes

47. YIELD CALCULATIONS PERCENT YIELD is the actual yield divided by the theoretical yield multiplied by 100. (By the way--YIELD is how much of the product is obtained.) PERCENT YIELD is the actual yield divided by the theoretical yield multiplied by 100. (By the way--YIELD is how much of the product is obtained.)

48. YIELD CALCULATIONS Analogy… Analogy… You get a grade of 12 on a chemistry test. (this is your actual yield). You get a grade of 12 on a chemistry test. (this is your actual yield). Is this good or bad? Is this good or bad? If the total possible on the test was a 15 (theoretical yield) then it is pretty good. If the total possible on the test was a 15 (theoretical yield) then it is pretty good. If the total possible is 250, then it is not so good. If the total possible is 250, then it is not so good.

49. YIELD CALCULATIONS The % score on the exam, (12/15)X100 = 80% is the percent yield--this is OK The % score on the exam, (12/15)X100 = 80% is the percent yield--this is OK The % score on the exam, (12/250)X100 = 5% is not so good. This percent yield means that additional work might be recommended. The % score on the exam, (12/250)X100 = 5% is not so good. This percent yield means that additional work might be recommended.

50. YIELD CALCULATIONS Let’s say that a student goes into the lab and mixes 25 grams of NaOH and 25 grams of HCl together. The student should obtain about 11 grams of NaCl--we determined that earlier. Let’s say that a student goes into the lab and mixes 25 grams of NaOH and 25 grams of HCl together. The student should obtain about 11 grams of NaCl--we determined that earlier. What if the student actually obtained 29 grams of NaCl? What if the student actually obtained 29 grams of NaCl? Which mass is “actual yield” and which is “theoretical yield”? Which mass is “actual yield” and which is “theoretical yield”?

51. YIELD CALCULATIONS The theoretical yield is 36.9 grams from previous calculations. The theoretical yield is 36.9 grams from previous calculations. The actual yield is 29 grams from the student’s experiment in the lab. The actual yield is 29 grams from the student’s experiment in the lab. The percent yield is The percent yield is (29/37) X 100 = 78% (29/37) X 100 = 78% This is OK, not great but it will work. This is OK, not great but it will work.

52. YIELD CALCULATIONS A second student went into the lab and did the same experiment. This student obtained a percent yield of 124%. A second student went into the lab and did the same experiment. This student obtained a percent yield of 124%. This student is REALLY GOOD! This student is REALLY GOOD! What was the actual yield of the student and how did they get more than 100%? What was the actual yield of the student and how did they get more than 100%?

53. YIELD CALCULATIONS The most likely reason is that the NaCl still contained some water and was not pure NaCl (contains impurities). Your lab partner dropping chewing gum in the beaker is not a valid reason.

54. Limiting Reactant to Percent Yield In the following synthesis reaction: 2 Na(s) + Cl 2 (g)  2 NaCl(s) If we have 53.2 grams of Na and 65.8 grams of Cl 2, what is the limiting reactant? Given: 53.2 grams Na 65.8 grams Cl 2 65.8 grams Cl 2 Find: Limiting Reactant Conversion: 2 moles Na  2 moles NaCl 1 mole Cl 2  2 moles NaCl

55. Limiting Reactant to Percent Yield

56. Add Another Step  Percent Yield Add Another Step  Percent Yield Percent Yield = Actual Yield X 100 = 86.4 grams X 100 = 80.0% Theoretical 108 grams Theoretical 108 grams

57. C 6 H 6 + HNO 3  C 6 H 5 NO 2 + H 2 0 Lets say in lab you react 20.3 grams of C 6 H 6 with excess nitric acid. a) Calculate the theoretical yield of C 6 H 5 NO 2. b) If you produced 28.7 grams of C 6 H 5 NO 2 in lab, what is your percentage yield? Another Percent Yield Problem….

60. Quantities in Chemical Reactions Chemical Principles Stoichiometry Stoichiometry Limiting Reactant, Theoretical Yield, and Percent Yield Limiting Reactant, Theoretical Yield, and Percent Yield

61. Quantities in Chemical Reactions Making Pancakes: Relationships Numerical Relationship Numerical Relationship 2 Eggs  5 Pancakes 2 Eggs  5 Pancakes Two eggs are required to produce 5 pancakes. Two eggs are required to produce 5 pancakes.

62. Stoichiometry Mg(s) + 2HCl(aq) → MgCl 2 (aq) + H 2 (g) If 2 mol of HCl react, how many moles of H 2 are obtained? If 2 mol of HCl react, how many moles of H 2 are obtained? How many moles of Mg will react with 2 mol of HCl? How many moles of Mg will react with 2 mol of HCl? If 4 mol of HCl react, how many mol of each product are produced? If 4 mol of HCl react, how many mol of each product are produced? How would you convert from moles of substances to masses? How would you convert from moles of substances to masses?

63. Quantities in Chemical Reactions Making Molecules: Mole to Mole Balanced Chemical Equation Balanced Chemical Equation Recipe Recipe 3 H 2 (g) + N 2 (g)  2 NH 3 (g) 3 H 2 (g) + N 2 (g)  2 NH 3 (g)

64. STOICHIOMETRY We have already done most of the TYPES of problems that we will see in this section. We have already done most of the TYPES of problems that we will see in this section. We are going to put several short problems together to make longer problems. We are going to put several short problems together to make longer problems.

65. STOICHIOMETRY Performing stoichiometry calculations is a 3 step process. These steps are: 1)convert what you are given into moles, 1)convert what you are given into moles, 2)use the mole to mole ratio from the BALANCED EQUATION to relate the two chemicals, and 2)use the mole to mole ratio from the BALANCED EQUATION to relate the two chemicals, and 3)convert the answer to the units requested in the problem (usually grams) 3)convert the answer to the units requested in the problem (usually grams)

66. STOICHIOMETRY The easiest way to learn stoichiometry is to do problems: The easiest way to learn stoichiometry is to do problems: How many grams of NaCl would be produced if 100 grams of NaOH react with excess HCl according to the following equation: NaOH + HCl  NaCl + H 2 O How many grams of NaCl would be produced if 100 grams of NaOH react with excess HCl according to the following equation: NaOH + HCl  NaCl + H 2 O

67. STOICHIOMETRY Before you even begin the problem, you should make sure the equation is balanced. Before you even begin the problem, you should make sure the equation is balanced. NaOH + HCl  NaCl + H 2 O NaOH + HCl  NaCl + H 2 O Yes, the equation is balanced. Yes, the equation is balanced.

68. STOICHIOMETRY Overall Look At The Steps Involved: Overall Look At The Steps Involved: NaOH + HCl  NaCl + H 2 O NaOH + HCl  NaCl + H 2 O

69. STOICHIOMETRY Step 1:Convert what you are given to moles. Step 1:Convert what you are given to moles. We are given 100 grams of NaOH at the beginning of the problem. We must convert this to moles: We are given 100 grams of NaOH at the beginning of the problem. We must convert this to moles:

70. STOICHIOMETRY Step 2:Use mole to mole ratio NaOH + HCl  NaCl + H 2 O Step 2:Use mole to mole ratio NaOH + HCl  NaCl + H 2 O We will use the ratio between NaCl (what the problem asks us to find) and the NaOH (what the problem gives us to start with) We will use the ratio between NaCl (what the problem asks us to find) and the NaOH (what the problem gives us to start with)

71. STOICHIOMETRY This ratio step looks like this--be sure to write all units and types of chemicals. This ratio step looks like this--be sure to write all units and types of chemicals.

72. STOICHIOMETRY Step 3)convert to the units asked for in the problem. Step 3)convert to the units asked for in the problem. The problem asks for number of grams of NaCl. So we need to convert the moles of NaCl to grams. The problem asks for number of grams of NaCl. So we need to convert the moles of NaCl to grams. 2.5 moles NaCl x 58.5 g/mole = 146 gNaCl 2.5 moles NaCl x 58.5 g/mole = 146 gNaCl Using significant figures, we get 150 g. Using significant figures, we get 150 g.

73. STOICHIOMETRY How about another problem? Using the equation given below, determine how many grams of Na 3 PO 4 would be needed to react with 5.5 grams of Ba(NO 3 ) 2 Ba(NO 3 ) 2 + Na 3 PO 4  Ba 3 (PO 4 ) 2 + NaNO 3 Ba(NO 3 ) 2 + Na 3 PO 4  Ba 3 (PO 4 ) 2 + NaNO 3

74. STOICHIOMETRY Although it is not one of the OFFICIAL steps, we should always make sure the equation is balanced BEFORE we being the problem. Remember, stoichiometry is the relationship between two chemicals in a BALANCED equation. Although it is not one of the OFFICIAL steps, we should always make sure the equation is balanced BEFORE we being the problem. Remember, stoichiometry is the relationship between two chemicals in a BALANCED equation. 3Ba(NO 3 ) 2 + 2Na 3 PO 4  Ba 3 (PO 4 ) 2 + 6 NaNO 3 3Ba(NO 3 ) 2 + 2Na 3 PO 4  Ba 3 (PO 4 ) 2 + 6 NaNO 3

75. 3Ba(NO 3 ) 2 + 2Na 3 PO 4  1Ba 3 (PO 4 ) 2 + 6 NaNO 3 Step 1) Convert 5.5 grams Ba(NO 3 ) 2 to moles Step 1) Convert 5.5 grams Ba(NO 3 ) 2 to moles Step 2) Convert moles of Ba(NO 3 ) 2 to moles of Step 2) Convert moles of Ba(NO 3 ) 2 to moles of Na 3 PO 4 (using the coefficients) Step 3) Convert moles of Na 3 PO 4 to grams of Na 3 PO 4

76. STOICHIOMETRY Sometimes you may not have to do all three steps--when you READ the problem it will give you clues. Sometimes you may not have to do all three steps--when you READ the problem it will give you clues. How many moles of Hydrogen gas will be produced when 2 moles of sodium react according to the following balanced equation? How many moles of Hydrogen gas will be produced when 2 moles of sodium react according to the following balanced equation?

77. 2Na + 2H 2 O  H 2 + 2NaOH The problem gives us some clues such as telling us that the equation is balanced. The problem gives us some clues such as telling us that the equation is balanced. Step 1) Convert to moles: Step 1) Convert to moles: We are given 2 moles of sodium so this step is not necessary. We are given 2 moles of sodium so this step is not necessary. Step 2) Use mole ratio from balanced equation: Step 2) Use mole ratio from balanced equation: Using the balanced equation… Using the balanced equation… 2 moles X 1 mole H 2 = 1 mole H 2 2 moles X 1 mole H 2 = 1 mole H 2 2 mole Na 2 mole Na Step 3) Convert to units requested in problem: Step 3) Convert to units requested in problem: The problem asked for moles of hydrogen gas--we already have that. The answer is 1 mole of hydrogen gas. The problem asked for moles of hydrogen gas--we already have that. The answer is 1 mole of hydrogen gas.

78. STOICHIOMETRY There are a few key points to remember about Stoichiometry. There are a few key points to remember about Stoichiometry. Stoichiometry can only look at TWO chemicals at one time Stoichiometry can only look at TWO chemicals at one time It does not matter if the two chemicals are It does not matter if the two chemicals are both products, both products, both reactants, both reactants, or one of each. or one of each. Stoichiometry problems can get harder than the examples. Stoichiometry problems can get harder than the examples.

79. LIMITING REACTANT The Limiting Reactant is defined as the reactant that runs out first in a chemical reaction. The Limiting Reactant is defined as the reactant that runs out first in a chemical reaction. 1 cup of Flour + 2 Eggs + ½ tsp Baking Powder  1 cup of Flour + 2 Eggs + ½ tsp Baking Powder  5 pancakes 5 pancakes 3 cups of Flour, 10 eggs, and 4 tsp baking powder 3 cups of Flour, 10 eggs, and 4 tsp baking powder 3 cups of flour  15 pancakes 3 cups of flour  15 pancakes 10 eggs  25 pancakes 10 eggs  25 pancakes 4 tsp baking powder  40 pancakes 4 tsp baking powder  40 pancakes

80. LIMITING REACTANT 1 cup of Flour + 2 Eggs + ½ tsp Baking Powder  5 pancakes 1 cup of Flour + 2 Eggs + ½ tsp Baking Powder  5 pancakes 3 cups of flour, 10 eggs, and 4 tsp baking powder 3 cups of flour, 10 eggs, and 4 tsp baking powder 3 cups of flour  15 pancakes 3 cups of flour  15 pancakes 10 eggs  25 pancakes 10 eggs  25 pancakes 4 tsp baking powder  40 pancakes 4 tsp baking powder  40 pancakes

81. LIMITING REACTANT 1 cup of Flour + 2 Eggs + ½ tsp Baking Powder  5 pancakes 1 cup of Flour + 2 Eggs + ½ tsp Baking Powder  5 pancakes 3 cups of flour  15 pancakes 3 cups of flour  15 pancakes Limiting Reactant (Reagent) Limiting Reactant (Reagent) Produces the Least Amount of Product Produces the Least Amount of Product

82. Limiting Reactant Example: Example: If we begin with 1.8 moles of titanium and 3.2 moles of chlorine, what is the limiting reactant in moles of TiCl 4 ? If we begin with 1.8 moles of titanium and 3.2 moles of chlorine, what is the limiting reactant in moles of TiCl 4 ? Ti(s) + 2 Cl 2 (g)  TiCl 4 (s) Ti(s) + 2 Cl 2 (g)  TiCl 4 (s) Given: 1.8 moles Ti and 3.2 moles Cl 2 Given: 1.8 moles Ti and 3.2 moles Cl 2 Find: Limiting reactant Find: Limiting reactant Conversions: Conversions: 1 mole Ti  1 mole TiCl 4 1 mole Ti  1 mole TiCl 4 2 mole Cl 2  1 mole TiCl 4 2 mole Cl 2  1 mole TiCl 4

83. Limiting Reactant If we begin with 1.8 moles of titanium and 3.2 moles of chlorine, what is the limiting reactant in moles of TiCl 4 ? (We are given information about TWO reactants and asked to find ONE of the products.) If we begin with 1.8 moles of titanium and 3.2 moles of chlorine, what is the limiting reactant in moles of TiCl 4 ? (We are given information about TWO reactants and asked to find ONE of the products.) Ti(s) + 2 Cl 2 (g)  TiCl 4 (s) Ti(s) + 2 Cl 2 (g)  TiCl 4 (s) 1.8 moles X 1 mole TiCl 4 = 1.8 moles TiCl 4 1.8 moles X 1 mole TiCl 4 = 1.8 moles TiCl 4 1 mole Ti 1 mole Ti 3.2 moles X 1 mole TiCl 4 = 1.6 moles TiCl 4 3.2 moles X 1 mole TiCl 4 = 1.6 moles TiCl 4 2 mole Cl 2 2 mole Cl 2

84. LIMITING REACTANT – Using grams Example: Example: 25 grams of HCl are mixed with 25 grams of NaOH. How many grams of NaCl can be made based on the following balanced chemical equation: 25 grams of HCl are mixed with 25 grams of NaOH. How many grams of NaCl can be made based on the following balanced chemical equation: HCl + NaOH  NaCl + H 2 O HCl + NaOH  NaCl + H 2 O

85. LIMITING REACTANT – Using grams Limiting reactant problems are not any harder than normal stoichiometry problems, they are just twice as long. We need to do TWO stoichiometry problems. Limiting reactant problems are not any harder than normal stoichiometry problems, they are just twice as long. We need to do TWO stoichiometry problems. The first problem will relate the NaOH to the NaCl. The first problem will relate the NaOH to the NaCl. The second problem will relate the HCl to the NaCl The second problem will relate the HCl to the NaCl

86. LIMITING REACTANT – Using grams After we have finished the two problems, we will obviously have two answers. After we have finished the two problems, we will obviously have two answers. The SMALLER of the two answers will be the limiting amount of NaCl that is produced and the reactant that produces the smaller number will be the limiting reactant (it will run out first). The SMALLER of the two answers will be the limiting amount of NaCl that is produced and the reactant that produces the smaller number will be the limiting reactant (it will run out first).

87. HCl + NaOH  NaCl + H 2 O PROBLEM #1 Step 1--convert 25 grams of NaOH to moles… Step 1--convert 25 grams of NaOH to moles… 25 grams x 1 mole/40 grams = 0.63 moles 25 grams x 1 mole/40 grams = 0.63 moles Step 2--use mole to mole ratio…. Step 2--use mole to mole ratio…. 0.63 moles NaOH x 1 mole NaCl = 0.63 moles of NaCl 0.63 moles NaOH x 1 mole NaCl = 0.63 moles of NaCl 1 mole NaOH 1 mole NaOH Step 3--convert to grams… Step 3--convert to grams… 0.63 moles NaCl x 58.5 g/mole = 36.9 grams of NaCl 0.63 moles NaCl x 58.5 g/mole = 36.9 grams of NaCl This is our first answer. This is our first answer. Now we can go to the second problem Now we can go to the second problem

88. HCl + NaOH  NaCl + H 2 O PROBLEM #2 Step 1--convert 25 grams HCl to moles… Step 1--convert 25 grams HCl to moles… 25 grams x 1 mole = 0.68 moles of HCl 25 grams x 1 mole = 0.68 moles of HCl 36.5 grams 36.5 grams Step 2--use mole ratio… Step 2--use mole ratio… 0.68 moles HCl x 1 mole NaCl = 0.68 moles NaCl 0.68 moles HCl x 1 mole NaCl = 0.68 moles NaCl 1 mole HCl 1 mole HCl Step 3--convert to grams… Step 3--convert to grams… 0.68 moles x 58.5g/mole = 39.8 g NaCl 0.68 moles x 58.5g/mole = 39.8 g NaCl Now we have our second answer. Now we have our second answer.

89. HCl + NaOH  NaCl + H 2 O The First Problem, looking at the relationship between NaOH and NaCl, gave us an answer of 36.9 grams of NaCl The First Problem, looking at the relationship between NaOH and NaCl, gave us an answer of 36.9 grams of NaCl The Second Problem, looking at the relationship between HCl and NaCl gave us an answer of 39.8 grams of NaCl The Second Problem, looking at the relationship between HCl and NaCl gave us an answer of 39.8 grams of NaCl

90. HCl + NaOH  NaCl + H 2 O The NaOH is the limiting reactant because it produced the smaller answer. The NaOH is the limiting reactant because it produced the smaller answer. The amount of NaCl that was produced in the reaction was 36.9 grams. The amount of NaCl that was produced in the reaction was 36.9 grams. Wasn’t that fun? Wasn’t that fun? But wait, we forgot to look at the fine print at the bottom of the problem. But wait, we forgot to look at the fine print at the bottom of the problem.

91. HCl + NaOH  NaCl + H 2 O Written in invisible power point the problem also said-- Written in invisible power point the problem also said-- How many grams of the NON-Limiting reactant will be left over at the end of the problem. How many grams of the NON-Limiting reactant will be left over at the end of the problem. HCl is the NON-limiting reactant. HCl is the NON-limiting reactant. Now we are back to a basic stoichiometry problem…. Now we are back to a basic stoichiometry problem….

92. HCl + NaOH  NaCl + H 2 O We can determine how much HCl was CONSUMED reacting with the 25 grams of NaOH. We can determine how much HCl was CONSUMED reacting with the 25 grams of NaOH. Once we know how much was consumed, we can subtract this from the 25 grams we started with to determine how much was left over. Once we know how much was consumed, we can subtract this from the 25 grams we started with to determine how much was left over.

93. HCl + NaOH  NaCl + H 2 O This is a simple three-part stoichiometry problem. This is a simple three-part stoichiometry problem. Step 1--convert to grams Step 1--convert to grams 25grams NaOH x 1 mole/40grams = 0.63mole of NaOH 25grams NaOH x 1 mole/40grams = 0.63mole of NaOH Step 2--mole to mole ratio Step 2--mole to mole ratio 0.63moles NaOH x 1mole HCl/1mole NaOH = 0.63 moles of HCl 0.63moles NaOH x 1mole HCl/1mole NaOH = 0.63 moles of HCl Step 3--convert to grams Step 3--convert to grams 0.63 moles X 36.5 g/mole = 23 grams of HCl 0.63 moles X 36.5 g/mole = 23 grams of HCl This HCl was CONSUMED This HCl was CONSUMED

94. HCl + NaOH  NaCl + H 2 O Since we started with 25 grams of HCL and we used 23 grams of HCl, there will be 2 grams of HCl leftover. Since we started with 25 grams of HCL and we used 23 grams of HCl, there will be 2 grams of HCl leftover. If we continue reading the invisible fine power point, the next part of the question states: If we continue reading the invisible fine power point, the next part of the question states: How many grams of the limiting reactant are left over? How many grams of the limiting reactant are left over?

95. HCl + NaOH  NaCl + H 2 O This is an easy stoichiometry problem. This is an easy stoichiometry problem. The answer is 0 grams. The answer is 0 grams. The limiting reactant is the one that runs out first therefore it can’t have leftovers. The limiting reactant is the one that runs out first therefore it can’t have leftovers. Be sure to read the problem and answer the problem the teacher asks and not the one that you think the teacher is going to ask. Be sure to read the problem and answer the problem the teacher asks and not the one that you think the teacher is going to ask.

96. HCl + NaOH  NaCl + H 2 O The last invisible power point question is: The last invisible power point question is: How many grams of water will be produced? How many grams of water will be produced? There are four ways to do this problem. There are four ways to do this problem. 1)relate HCl to H 2 O 1)relate HCl to H 2 O 2)relate NaOH to H 2 O 2)relate NaOH to H 2 O 3)relate NaCl to H 2 O 3)relate NaCl to H 2 O 4)use the Law of Conservation of Mass. 4)use the Law of Conservation of Mass.

97. HCl + NaOH  NaCl + H 2 O I am sure that you will be surprised to learn that we are going to do this last problem four different ways and see if we get the same answer. I am sure that you will be surprised to learn that we are going to do this last problem four different ways and see if we get the same answer.

98. HCl + NaOH  NaCl + H 2 O What do we know from previous problems? What do we know from previous problems? 25 grams of NaOH is the limiting reactant 25 grams of NaOH is the limiting reactant 23 grams of HCl was also used 23 grams of HCl was also used 36.9 grams of NaCl was produced. 36.9 grams of NaCl was produced. We will do the problems with a minimum of discussion to save time and space. We will do the problems with a minimum of discussion to save time and space.

99. HCl + NaOH  NaCl + H 2 O Calculations from 23grams of HCl: Calculations from 23grams of HCl:

100. HCl + NaOH  NaCl + H 2 O Law of Conservation of Mass We used 23 grams of HCl and 25 grams of NaOH. Therefore, the total mass of the reactants that was used was 48 grams. We used 23 grams of HCl and 25 grams of NaOH. Therefore, the total mass of the reactants that was used was 48 grams. The Law of Conservation of Mass tells us that what we start with is what we end with. The Law of Conservation of Mass tells us that what we start with is what we end with. Therefore, we must end with 48 grams of products. We ended with 36.9 grams of NaCl, so we must have had 11.1 grams of H 2 O based on Conservation of Mass. Therefore, we must end with 48 grams of products. We ended with 36.9 grams of NaCl, so we must have had 11.1 grams of H 2 O based on Conservation of Mass. (Remember--the other two grams of HCl that we DID NOT use does not count either way.) (Remember--the other two grams of HCl that we DID NOT use does not count either way.)

101. HCl + NaOH NaCl + H 2 O We now have four answers: Starting with HCl we ended up with 11.3 grams Starting with HCl we ended up with 11.3 grams Starting with NaOH we ended up with 11.3 grams Starting with NaOH we ended up with 11.3 grams Starting with NaCl we ended up with 11.3 grams, Starting with NaCl we ended up with 11.3 grams, Using the Law of Conservation of Mass, we ended up with 11.1 grams. Using the Law of Conservation of Mass, we ended up with 11.1 grams. They are different--but not much! They are different--but not much!

102. HCl + NaOH NaCl + H 2 O The reason for the difference is the rounding that took place during the problem. The reason for the difference is the rounding that took place during the problem. If you remember sig figs, we should only have 2 sig figs anyway based on the 25 grams of starting material. If you remember sig figs, we should only have 2 sig figs anyway based on the 25 grams of starting material. Using correct sig fig’s, all methods give the same answer = 11 grams of water. Using correct sig fig’s, all methods give the same answer = 11 grams of water.

103. YIELD CALCULATIONS Let’s go to another topic... Let’s go to another topic... THEORETICAL YIELD is the mathematical answer produced based solely on calculations. THEORETICAL YIELD is the mathematical answer produced based solely on calculations.

104. More Pancakes: Limiting Reactant, Theoretical Yield, and Percent Yield 1 cup of Flour + 2 Eggs + ½ tsp Baking Powder  5 pancakes 1 cup of Flour + 2 Eggs + ½ tsp Baking Powder  5 pancakes Theoretical Yield Theoretical Yield 15 pancakes  Most that can be produced 15 pancakes  Most that can be produced Actual Yield Actual Yield What was actually produced  11 pancakes What was actually produced  11 pancakes Burned 3 pancakes Burned 3 pancakes Dropped 1 pancake Dropped 1 pancake Percent Yield Percent Yield Percent of the theoretical yield actually attained Percent of the theoretical yield actually attained Percent Yield Percent Yield Percent of the theoretical yield actually attained Percent of the theoretical yield actually attained = 11 pancakes X 100 = 73% 15 pancakes 15 pancakes

105. YIELD CALCULATIONS PERCENT YIELD is the actual yield divided by the theoretical yield multiplied by 100. (By the way--YIELD is how much of the product is obtained.) PERCENT YIELD is the actual yield divided by the theoretical yield multiplied by 100. (By the way--YIELD is how much of the product is obtained.)

106. YIELD CALCULATIONS Analogy… Analogy… You get a grade of 12 on a chemistry test. (this is your actual yield). You get a grade of 12 on a chemistry test. (this is your actual yield). Is this good or bad? Is this good or bad? If the total possible on the test was a 15 (theoretical yield) then it is pretty good. If the total possible on the test was a 15 (theoretical yield) then it is pretty good. If the total possible is 250, then it is not so good. If the total possible is 250, then it is not so good.

107. YIELD CALCULATIONS The % score on the exam, (12/15)X100 = 80% is the percent yield--this is OK The % score on the exam, (12/15)X100 = 80% is the percent yield--this is OK The % score on the exam, (12/250)X100 = 5% is not so good. This percent yield means that additional work might be recommended. The % score on the exam, (12/250)X100 = 5% is not so good. This percent yield means that additional work might be recommended.

108. YIELD CALCULATIONS Let’s say that a student goes into the lab and mixes 25 grams of NaOH and 25 grams of HCl together. The student should obtain about 11 grams of NaCl--we determined that earlier. Let’s say that a student goes into the lab and mixes 25 grams of NaOH and 25 grams of HCl together. The student should obtain about 11 grams of NaCl--we determined that earlier. What if the student actually obtained 9 grams of NaCl? What if the student actually obtained 9 grams of NaCl? Which mass is “actual yield” and which is “theoretical yield”? Which mass is “actual yield” and which is “theoretical yield”?

109. YIELD CALCULATIONS The theoretical yield is 11 grams from previous calculations. The theoretical yield is 11 grams from previous calculations. The actual yield is 9 grams from the student’s experiment in the lab. The actual yield is 9 grams from the student’s experiment in the lab. The percent yield is The percent yield is (9/11) X 100 = 82% (9/11) X 100 = 82% This is OK, not great but it will work. This is OK, not great but it will work.

110. YIELD CALCULATIONS A second student went into the lab and did the same experiment. This student obtained a percent yield of 124%. A second student went into the lab and did the same experiment. This student obtained a percent yield of 124%. This student is REALLY GOOD! This student is REALLY GOOD! What was the actual yield of the student and how did they get more than 100%? What was the actual yield of the student and how did they get more than 100%?

111. YIELD CALCULATIONS Actual yield divided by theoretical yield times 100 = percent yield. Actual yield divided by theoretical yield times 100 = percent yield. Let’s call the actual yield “Z”--a neat name for an unknown borrowed from algebra class. Let’s call the actual yield “Z”--a neat name for an unknown borrowed from algebra class. (Z/11) X 100 = 124% (Z/11) X 100 = 124% Therefore, Z = (124)(11)/100 =14 grams Therefore, Z = (124)(11)/100 =14 grams

112. YIELD CALCULATIONS The actual yield for the student was 14 grams. The actual yield for the student was 14 grams. How did the student get more than the 11 grams theoretically possible? How did the student get more than the 11 grams theoretically possible? The most likely reason is that the NaCl still contained some water and was not pure NaCl (contains impurities). The most likely reason is that the NaCl still contained some water and was not pure NaCl (contains impurities). Your lab partner dropping chewing gum in the beaker is not a valid reason. Your lab partner dropping chewing gum in the beaker is not a valid reason.

113. Limiting Reactant to Percent Yield In the following synthesis reaction: 2 Na(s) + Cl 2 (g)  2 NaCl(s) 2 Na(s) + Cl 2 (g)  2 NaCl(s) If we have 53.2 grams of Na and 65.8 grams of Cl 2, what is the limiting reactant? If we have 53.2 grams of Na and 65.8 grams of Cl 2, what is the limiting reactant? Given: 53.2 grams Na Given: 53.2 grams Na 65.8 grams Cl 2 65.8 grams Cl 2 Find: Limiting Reactant Find: Limiting Reactant Conversion: Conversion: 2 moles Na  2 moles NaCl 2 moles Na  2 moles NaCl 1 mole Cl 2  2 moles NaCl 1 mole Cl 2  2 moles NaCl

114. Limiting Reactant to Percent Yield Add Another Step  Percent Yield Add Another Step  Percent Yield Percent Yield = Actual Yield X 100 = 86.4 grams X 100 = 80.0% Percent Yield = Actual Yield X 100 = 86.4 grams X 100 = 80.0% Theoretical 108 grams Theoretical 108 grams

115. C 6 H 6 + HNO 3  C 6 H 5 NO 2 + H 2 0 Lets say in lab you react 20.3 grams of C 6 H 6 with excess nitric acid. Lets say in lab you react 20.3 grams of C 6 H 6 with excess nitric acid. a) Calculate the theoretical yield of C 6 H 5 NO 2. a) Calculate the theoretical yield of C 6 H 5 NO 2. b) If you produced 28.7 grams of C 6 H 5 NO 2 in lab, what is your percentage yield? b) If you produced 28.7 grams of C 6 H 5 NO 2 in lab, what is your percentage yield?

116. Quantities in Chemical Reactions Chemical Principles Stoichiometry Stoichiometry Limiting Reactant, Theoretical Yield, and Percent Yield Limiting Reactant, Theoretical Yield, and Percent Yield

117. Quantities in Chemical Reactions Chemical Skills Mole to Mole Conversions Mole to Mole Conversions Mass to Mass Conversions Mass to Mass Conversions Limiting Reactant, Theoretical Yield, and Percent Yield Limiting Reactant, Theoretical Yield, and Percent Yield