Presentation is loading. Please wait.

Presentation is loading. Please wait.

% composition problems and combustion analysis (pp. 94-103) Reaction balancing (pp. 105-108) …cont. Reaction stoichiometry predictions, limiting yields.

Published byAllan Kelley Farmer Modified over 9 years ago

Similar presentations

Presentation on theme: "% composition problems and combustion analysis (pp. 94-103) Reaction balancing (pp. 105-108) …cont. Reaction stoichiometry predictions, limiting yields."— Presentation transcript:

1 % composition problems and combustion analysis (pp. 94-103) Reaction balancing (pp. 105-108) …cont. Reaction stoichiometry predictions, limiting yields and % yields ( pp. 108-123) The mole road trip itinerary...

2 How to Reaction balance (pp. 105-108) Essentially, just try to make elements balance by trial and error One more `trickier’ one

3 More in-class equation balancing practice __ H 3 PO 3  ___H 3 PO 4 + ___PH 3 4 3 1 Helpful hint: first balance elements that appear in a single compound on both reactant and product sides.

4 ____C 6 H 14 + ___O 2  ___CO 2 +___H 2 O Balance me, please !!! …On Board ( a trickier one…) 2 12 14 13

5 Reaction Stoichiometry problems: Moles level 3 how chemists cook

6 Reaction stoichiometry means following….. THE RECIPE

7 A simple warm-up with kitchen chemistry

8 2+ 3 + 300 g/block 500 g/dozen 200 g/box 1 1900 g/souffle How many boxes of cream to make 4 souffles ? 12 THE RECIPE…

9 Cooking and stoichiometry 2+ 3 + 300 g/block 500 g/dozen 200 g/box 1 1900 g/souffle 600 grams cheese makes how many souffles ? THE RECIPE 2

10 Cooking and stoichiometry 2+ 3 + 300 g/block 500 g/dozen 200 g/box 1 1900 g/souffle How many grams of eggs combine with 9 boxes of cream ? THE RECIPE 3000 g

11 Cooking and stoichiometry 2+ 3 + 300 g/block 500 g/dozen 200 g/box 1 1900 g/souffle 48 eggs combine with how many blocks of cheese? THE RECIPE 2

12 How many moles of O 2 will burn to form 1.2 moles of CO 2 ? Simple mol-mol stoichiometry conversions C 3 H 8 + 5O 2 --------  3CO 2 + 4H 2 O Method 1: factor label way 1.2 mol CO 2 * 5 mol O 2 3 mol CO 2 1) given 3) Use reaction stoichiometry coefficients in right ratio to cancel and connect…which ratio ??? = ? mol O 2 2) want 2

13 How many moles of O 2 will burn to form 1.2 moles of CO 2 ? C 3 H 8 + 5O 2 --------  3CO 2 + 4H 2 O 5 3 1.2* m= 1.2 *5 1.2 3 3)Solve for m 1.2 mol ?? m Problem stated visually m 1.2 Method 2: `mole ratios’ way Mol-mol stoichiometry conversions (continued) 1) ratio `wanted’ moles in numerator to given in denominator 2) Set equal to matching coefficients for compounds given in reaction…which ??? = = 2

14 More complex stoichiometry problem done 2 ways C 3 H 8 + 5O 2  3CO 2 + 4H 2 O 22 grams of C 3 H 8 burned with O 2 makes how many grams of H 2 O ? Method 1: factor label 22 g C 3 H 8 1 mole C 3 H 8 44 g C 3 H 8 4 mol H 2 O 1 mole C 3 H 8 MW44 3244 18 g/mol 18 g H 2 O 1 mol H 2 O = ?? g H 2 O 36 x x x

15 Method 2: mole ratio way (Board first) C 3 H 8 + 5O 2  3CO 2 + 4H 2 O 22 grams of C 3 H 8 burned with O 2 makes how many grams of H 2 O ? 44 3244 18 g/mol ?? g Problem stated visually Give n : 22 g

16 Method 2: mole ratio way C 3 H 8 + 5O 2  3CO 2 + 4H 2 O 22 grams of C 3 H 8 burned with O 2 makes how many grams of H 2 O ? 44 3244 18 g/mol ?? g Problem stated visually Give n : 22 g 0 ) convert all given masses & molecule counts to moles 22 g C 3 H 8 * 1 mol C 3 H 8 44 g C 3 H 8 = 0.5 mol C 3 H 8

17 Method 2: mole ratio way C 3 H 8 + 5O 2  3CO 2 + 4H 2 O 22 grams of C 3 H 8 burned with O 2 makes how many grams of H 2 O ? 44 3244 18 g/mol ?? g Problem stated visually Give n : 22 g 1) ratio `wanted’ moles m in numerator to given in denominator: = 0.5 mol C 3 H 8 m = mol H 2 O 0.5 mol C 3 H 8 = 4 1

18 Method 2: mole ratio way C 3 H 8 + 5O 2  3CO 2 + 4H 2 O 22 grams of C 3 H 8 burned with O 2 makes how many grams of H 2 O ? 44 3244 18 g/mol ?? g Give n : 22 g m 0.5 = 4 1 2) Solve for wanted moles 0.5 * *0.5= 2 mol H 2 O

19 Method 2: mole ratio way C 3 H 8 + 5O 2  3CO 2 + 4H 2 O 22 grams of C 3 H 8 burned with O 2 makes how many grams of H 2 O ? 44 3244 18 g/mol ?? g Give n : 22 g = 2 mol H 2 O 3) Convert wanted moles to wanted final units (grams) 2 mol H 2 O * 18 g H 2 O 1 mole H 2 O = 36 g H 2 O

20 Practice, practice, practice….

21 Sample Reaction 1 Chemical Stoichiometry Problems Solved in Detail Sample reaction 1 C 3 H 8 + 5O 2 ------ 3CO 2 + 4H 2 O 44 32 44 18g/mol First U try them…then we’ll work through sample reaction 1: problems a-f the mole ratio way

22 a) mol-mol: How many moles of O 2 will burn to form 1.60 moles of H 2 O? A. 0.8 B.1.28 C.2.00 D.0.20 C 3 H 8 + 5O 2 --------  3CO 2 + 4H 2 O (BOOM) 44 32 44 18g/mol

23 b) moles to weight How many grams of CO 2 are generated if 0.0757 moles of O 2 are burned? C 3 H 8 + 5O 2 --------  3CO 2 + 4H 2 O (BOOM) 44 32 44 18g/mol 2 g

24 c)wt-wt: How many grams of O 2 are needed to burn 1.1 g C 3 H 8 ? A.0.8 B.4.0 C.0.25 D.0.16 C 3 H 8 + 5O 2 --------  3CO 2 + 4H 2 O (BOOM) 44 32 44 18g/mol

25 d) weight to moles How many moles of H 2 O form if 33 g of C 3 H 8 are burned ? C 3 H 8 + 5O 2 --------  3CO 2 + 4H 2 O (BOOM) 44 32 44 18g/mol 3 mol H 2 O

26 e) How many grams of O 2 are needed to form 9*10 22 molecules of H 2 O ? A.1.0 g B.2.0 g C.1.5 g D.6.0 g C 3 H 8 + 5O 2 --------  3CO 2 + 4H 2 O (BOOM) 44 32 44 18g/mol

27 f ) weight to count How many molecules of CO 2 form if 1.592 g H 2 O results ? C 3 H 8 + 5O 2 --------  3CO 2 + 4H 2 O (BOOM) 44 32 44 18 g/mol =4*10 22

28 % composition problems and combustion analysis (pp. 94-103) Reaction balancing (pp. 105-108) Reaction stoichiometry predictions, limiting yields and % yields ( pp. 108-123) The mole road trip itinerary...

29 Limiting yield and % yield calculations are the last stop on the mole bus trip

30 A non-chemical example of a `limiting’ yield problem You have gotten lost on the Ad Dahna desert –largest desert on the Saudi peninisula. To avoid perishing you must reach the nearest oasis which is 100 km away. You can walk at maximum 10 km/ day. You need to consume at least 2 liters of water and ½ kg of food per day to walk that distance. You have in your pack: 16 liters of water 5 kg of food Broken cell phone What’s the maximum distance you can expect to travel ??

31 Food Calculation 5 kg food = 10 days ½ kg/day => 10 days * 10 km = 100 km day

32 16 liters = 8 days 2 liters/day Water Calculation  8 days * 10 km = 80 km  day The winner is….always the smaller one… it limits.

33 3 examples of limiting yield calculations via the `cut and try’ approach (on board) example #1 5O 2 + C 3 H 8  3CO 2 + 4H 2 O mol 0.33 0.05 ?? mol Given 0.33 mol O 2 and 0.10 mol C 3 H 8, compute the maximum theoretical yield of CO 2 moles Ans. 0.15 mol CO 2 (C 3 H 8 limits)

34 The `cut & try’ approach to limiting yield calculations (cont.) 5O 2 + C 3 H 8  3CO 2 + 4H 2 O MW 32 44 44 18 g 4.0 2.2 ? g example #2: 0.18 g H 2 O (O 2 limits) Given 4.0 grams O 2 and 2.2 grams C 3 H 8, compute the theoretical yield of H 2 O for the combustion shown above.

35 C 12 H 22 O 11 + 12O 2 --------  12CO 2 + 11H 2 O MW 342 g/mol 32 g/mol 44 g/mol 18 g/mol w(g) 68.40 72.73?? g How many grams of CO 2 will be produced if 68.40 grams of sucrose, C 12 H 22 O 11, is combined with 72.73 grams of O 2 according to the balanced equation above? 100 g CO 2 (O 2 limits) Cut and try approach example #3:

36 Guided practice limiting yield problems Mole-mol-mol a)0.25 moles of C 4 H 10 and 1.4 moles of O 2 are reacted. What is the maximum yield of CO 2 in moles? 2C 4 H 10 + 13O 2 --------  8CO 2 + 10H 2 O 58 32 44 18g/mol 0.25 mol C 4 H 10 produces 8*0.25/2 =1 mol CO 2 1.4 mol O 2 produces 8*0.25/13 =0.15 mol CO 2

37 Guided practice limiting yield problems (cont.) Weight-weight-mol b)1 gram of C 4 H 10 and 10 grams of O 2 are reacted. What is the maximum yield in H 2 O in moles ? 2C 4 H 10 + 13O 2 --------  8CO 2 + 10H 2 O 58 32 44 18g/mol 0.017 mol C 4 H 10 produces 10*0.017/2=0.085 mol H 2 O 0.312 mol O 2 produces 10*0.312/13 =0.24 mol H 2 O 1 gram C 4 H 10 = 1/58=0.017 mol C 4 H 10 10 gram O 2 = 10/32=0.312 mol O 2

38 Weight-weight-molecules c) 5.8 grams of C 4 H 10 and 160 grams of O 2 are reacted. What is the maximum yield of CO 2 in molecule count? 5.2 Calculate the maximum yield problems (cont.) 2C 4 H 10 + 13O 2 --------  8CO 2 + 10H 2 O 58 32 44 18g/mol 5.8/58=0.1 mol C 4 H 10 160/32=5 mol O 2 0.1 mol C 4 H 10 produces 8*0.1/2=0.4 mol CO 2 5 mol O 2 produces 8*5/13 =3.07 mol CO 2 C 4 H 10 limits 0.4*6*10 23 = molecules CO 2 = 2.4*10 23

39 5.2 Calculate the maximum yield problems (cont.) 2C 4 H 10 + 13O 2 --------  8CO 2 + 10H 2 O 58 32 44 18g/mol Weight-molecules-weight d)116 grams of C 4 H 10 and 1.66*10 24 molecules of O 2 react. What is the maximum yield of H 2 O in grams ? 116/58=2 mol C 4 H 10 1.66*10 24 /6*10 23 =1.1 mol O 2 2 mol C 4 H 10 produces 10*2/2=10 mol H 2 O 1.1 mol O 2 produces 10*1.1/13 =0.85 mol H 2 O O 2 limitsMultiply down to calculate: grams H 2 O= 0.85*18=15.3 g

40 U-Try-It on your own Clicker Examples

41 2C 4 H 10 + 13O 2 --------  8CO 2 + 10H 2 O Given 0.5 mol of C 4 H 10 and 6.5 mol of O 2, how many mol of CO 2 will form? A. 8 B. 4 C. 2 D. 1

42 2C 4 H 10 + 13O 2 --------  8CO 2 + 10H 2 O 58324418 g/mol Given 10 g of C 4 H 10 and 100 g of O 2, how many mol of H 2 O will form? A. 2.4 B. 0.17 C. 0.86 D. 3.12 E.0.034

43 2C 4 H 10 + 13O 2 --------  8CO 2 + 10H 2 O 58324418 g/mol Given 2*10 22 molecules of C 4 H 10 and 50 g of O 2, how many g of CO 2 will form? 1 mol count =6*10 23 A. 5.8 g B. 0.133 g C. 4.22 g D. 1.2 g

Download ppt "% composition problems and combustion analysis (pp. 94-103) Reaction balancing (pp. 105-108) …cont. Reaction stoichiometry predictions, limiting yields."

Similar presentations

Stoichiometry: The study of quantitative measurements in chemical formulas and reactions Chemistry – Mrs. Cameron.

Stoichiometry: The study of quantitative measurements in chemical formulas and reactions Chemistry – Mrs. Cameron.
Stoichiometry (Yay!).

Stoichiometry (Yay!).
Chapter 3: Calculations with Chemical Formulas and Equations MASS AND MOLES OF SUBSTANCE 3.1 MOLECULAR WEIGHT AND FORMULA WEIGHT -Molecular weight: (MW)

Chapter 3: Calculations with Chemical Formulas and Equations MASS AND MOLES OF SUBSTANCE 3.1 MOLECULAR WEIGHT AND FORMULA WEIGHT -Molecular weight: (MW)
/11 Multistep Conversions Make 20 cookies!

/11 Multistep Conversions Make 20 cookies!
Stoichiometry Applying equations to calculating quantities.

Stoichiometry Applying equations to calculating quantities.
Chemical Quantities In Reactions

Chemical Quantities In Reactions
Calculations with Chemical Formulas and Equations

Calculations with Chemical Formulas and Equations
Chapter 41 Chemical Equations and Stoichiometry Chapter 4.

Chapter 41 Chemical Equations and Stoichiometry Chapter 4.
Chemical Reactions Unit

Chemical Reactions Unit
Chapter 9 Chemical Quantities. 9 | 2 Information Given by the Chemical Equation Balanced equations show the relationship between the relative numbers.

Chapter 9 Chemical Quantities. 9 | 2 Information Given by the Chemical Equation Balanced equations show the relationship between the relative numbers.
Chapter 3.  Reactants are left of the arrow  Products are right of the arrow  The symbol  is placed above the arrow to indicate that the rxn is being.

Chapter 3.  Reactants are left of the arrow  Products are right of the arrow  The symbol  is placed above the arrow to indicate that the rxn is being.
Chemistry 1210: General Chemistry Dr. Gina M. Florio 13 September 2012 Jespersen, Brady, Hyslop, Chapter 4 The Mole and Stoichiometry.

Chemistry 1210: General Chemistry Dr. Gina M. Florio 13 September 2012 Jespersen, Brady, Hyslop, Chapter 4 The Mole and Stoichiometry.
Chapter 3 Stoichiometry Stoichiometry: The study of quantities of materials consumed and produced in chemical reactions Atomic Masses: Are determined by.

Chapter 3 Stoichiometry Stoichiometry: The study of quantities of materials consumed and produced in chemical reactions Atomic Masses: Are determined by.
The Mole Atomic mass provides a means to count atoms by measuring the mass of a sample The periodic table on the inside cover of the text gives atomic.

The Mole Atomic mass provides a means to count atoms by measuring the mass of a sample The periodic table on the inside cover of the text gives atomic.
Chapter 2-6 – 2-12 Chapter 3-1 – 3-8

Chapter 2-6 – 2-12 Chapter 3-1 – 3-8
Stoichiometry: the mass relationships between reactants and products. We will use the molar masses ( amount of grams in one mole of a element or compound)

Stoichiometry: the mass relationships between reactants and products. We will use the molar masses ( amount of grams in one mole of a element or compound)
April 3, 2014 Stoichiometry. Stoichiometry is the study of quantities of materials consumed and produced in chemical reactions Stoikheion (Greek, “element”)

April 3, 2014 Stoichiometry. Stoichiometry is the study of quantities of materials consumed and produced in chemical reactions Stoikheion (Greek, “element”)
Aim: Using mole ratios in balanced chemical equations.

Aim: Using mole ratios in balanced chemical equations.
Calculations with Chemical Formulas and Equations.

Calculations with Chemical Formulas and Equations.
Chapter 3. Atoms are very small, but we need to know the mass of different atoms to compare them. To do this, we define a unit, called the atomic mass.

Chapter 3. Atoms are very small, but we need to know the mass of different atoms to compare them. To do this, we define a unit, called the atomic mass.

Similar presentations

Ads by Google