% composition problems and combustion analysis (pp. 94-103) Reaction balancing (pp. 105-108)...

transcript

• % 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...

How to Reaction balance (pp. 105-108)

Essentially, just try to make elements balance by trial and error

One more `trickier’ one

More in-class equation balancing practice

__H3PO3 ___H3PO4 + ___PH34 3 1

Helpful hint: first balance elements that appear in a single compound on both reactant and product sides.

____C6H14 + ___O2 ___CO2 +___H2O

Balance me, please !!! …On Board( a trickier one…)

2 12 1413

Reaction Stoichiometry problems: Moles level 3how chemists cook

Reaction stoichiometry means following…..THE RECIPE

A simple warm-up with kitchen chemistry

300 g/block 500 g/dozen 200 g/box

1900 g/souffle

• How many boxes of cream to make 4 souffles ?

THE RECIPE…

Cooking and stoichiometry

1900 g/souffle

• 600 grams cheese makes how many souffles ?

THE RECIPE

1900 g/souffle

• How many grams of eggs combine with 9 boxes of cream ?

THE RECIPE

3000 g

1900 g/souffle

• 48 eggs combine with how many blocks of cheese?

THE RECIPE

How many moles of O2 will burn to form 1.2 moles of CO2 ?

Simple mol-mol stoichiometry conversionsC3H8 + 5O2-------- 3CO2 + 4H2O

Method 1: factor label way

1.2 mol CO2 * 5 mol O2

3 mol CO2

1) given

3) Use reaction stoichiometry coefficients in right ratio to cancel and connect…which ratio ???

= ? mol O2

2) want

How many moles of O2 will burn to form 1.2 moles of CO2 ?

C3H8 + 5O2-------- 3CO2 + 4H2O

1.2* m= 1.2 *5 1.2 3

3)Solve for m

1.2 mol?? mProblem stated visually

Method 2: `mole ratios’ wayMol-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 ???

More complex stoichiometry problem done 2 ways

C3H8+ 5O2 3CO2+ 4H2O22 grams of C3H8 burned with O2 makes how

many grams of H2O ?Method 1: factor label

22 g C3H8 1 mole C3H8

44 g C3H8

4 mol H2O1 mole C3H8

MW 44 32 44 18 g/mol

18 g H2O 1 mol H2O

= ?? g H2O36

Method 2: mole ratio way (Board first)

C3H8+ 5O2 3CO2+ 4H2O

22 grams of C3H8 burned with O2 makes how many grams of H2O ?

44 32 44 18 g/mol

Problem statedvisually

Given:

Method 2: mole ratio way

C3H8+ 5O2 3CO2+ 4H2O

44 32 44 18 g/mol

Given:

22 g0) convert all given masses & molecule counts to moles

22 g C3H8 * 1 mol C3H8

44 g C3H8

= 0.5 mol C3H8

C3H8+ 5O2 3CO2+ 4H2O

44 32 44 18 g/mol

Given:

1) ratio `wanted’ moles m in numerator to given in denominator:

= 0.5 mol C3H8

m = mol H2O0.5 mol C3H8

C3H8+ 5O2 3CO2+ 4H2O

44 32 44 18 g/mol

Given:

2) Solve for wanted moles0.5 * *0.5 = 2 mol H2O

C3H8+ 5O2 3CO2+ 4H2O

44 32 44 18 g/mol

Given:

22 g= 2 mol H2O

3) Convert wanted moles to wanted final units (grams)

2 mol H2O * 18 g H2O 1 mole H2O

= 36 g H2O

Practice, practice, practice….

Sample Reaction 1 Chemical Stoichiometry

Problems Solved in Detail

Sample reaction 1C3H8 + 5O2------ 3CO2 + 4H2O 44 32 44 18 g/mol

First U try them…then we’ll work through sample reaction 1: problems a-f the mole ratio way

a) mol-mol: How many moles of O2 will burn

to form 1.60 moles of H2O?

A. 0.8

B.1.28

C.2.00

D.0.20

26.40.36 1 0.2

C3H8 + 5O2-------- 3CO2 + 4H2O (BOOM) 44 32 44 18g/mol

b) moles to weight How many grams of CO2 are generated if 0.0757 moles of O2 are burned?

C3H8 + 5O2-------- 3CO2 + 4H2O (BOOM) 44 32 44 18 g/mol

c)wt-wt: How many grams of O2 are needed to burn 1.1 g C3H8?

C.0.25

D.0.16

0.0125 0.6

7% 9%7%

C3H8 + 5O2-------- 3CO2 + 4H2O (BOOM) 44 32 44 18g/mol

d) weight to moles How many moles of H2O form if 33 g of C3H8 are burned

3 mol H2O

e) How many grams of O2 are needed to

form 9*1022 molecules of H2O ?

A. 1.0 g

B. 2.0 g

C. 1.5 g

D. 6.0 g

1.0 g2.0 g

1.5 g6.0 g

f) weight to count How many molecules of CO2 form if

1.592 g H2O results ?

=4*1022

• % 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...

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

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 ??

Food Calculation

5 kg food = 10 days½ kg/day

=>10 days * 10 km = 100 km day

16 liters = 8 days2 liters/day

Water Calculation

8 days * 10 km = 80 km day

The winner is….always the smaller one…it limits.

3 examples of limiting yield calculationsvia the `cut and try’ approach (on board)

example #1

5O2 + C3H8 3CO2 + 4H2O mol 0.33 0.05 ?? mol

Given 0.33 mol O2 and 0.10 mol C3H8, compute the maximum theoretical yield of CO2 moles

Ans. 0.15 mol CO2 (C3H8 limits)

The `cut & try’ approach to limiting yield calculations (cont.)

5O2 + C3H8 3CO2 + 4H2O

MW 32 44 44 18 g 4.0 2.2 ? g

example #2: 0.18 g H2O

(O2 limits)

Given 4.0 grams O2 and 2.2 grams C3H8, compute the theoretical yield of H2O for the combustion shown above.

C12H22O11 + 12O2 -------- 12CO2 + 11H2O

MW 342 g/mol 32 g/mol 44 g/mol 18 g/molw(g) 68.40 72.73 ?? g

How many grams of CO2 will be produced if

68.40 grams of sucrose, C12H22O11, is

combined with 72.73 grams of O2 according

to the balanced equation above?

100 g CO2

(O2 limits)

Cut and try approach example #3:

Guided practice limiting yield problems

Mole-mol-mola)0.25 moles of C4H10 and 1.4 moles of O2 are reacted. What is the maximum yield of CO2 in moles?

2C4H10 + 13O2-------- 8CO2 + 10H2O 58 32 44 18 g/mol

0.25 mol C4H10 produces 8*0.25/2 =1 mol CO2

1.4 mol O2 produces 8*0.25/13 =0.15 mol CO2

Guided practice limiting yield problems (cont.)

Weight-weight-molb)1 gram of C4H10 and 10 grams of O2 are reacted. What is the maximum yield in H2O in moles ?

2C4H10 + 13O2-------- 8CO2 + 10H2O 58 32 44 18 g/mol

0.017 mol C4H10 produces 10*0.017/2=0.085 mol H2O

0.312 mol O2 produces 10*0.312/13 =0.24 mol H2O

1 gram C4H10 = 1/58=0.017 mol C4H10

10 gram O2 = 10/32=0.312 mol O2

Weight-weight-moleculesc) 5.8 grams of C4H10 and 160 grams of O2 are reacted. What is the maximum yield of CO2 in molecule count?

5.2 Calculate the maximum yield problems (cont.)2C4H10 + 13O2-------- 8CO2 + 10H2O 58 32 44 18 g/mol

5.8/58=0.1 mol C4H10 160/32=5 mol O2

0.1 mol C4H10 produces 8*0.1/2=0.4 mol CO2

5 mol O2 produces 8*5/13 =3.07 mol CO2

C4H10 limits

0.4*6*1023 = molecules CO2= 2.4*1023

5.2 Calculate the maximum yield problems (cont.)2C4H10 + 13O2-------- 8CO2 + 10H2O 58 32 44 18 g/molWeight-molecules-weightd)116 grams of C4H10 and 1.66*1024 molecules of O2 react. What is the maximum yield of H2O in grams?

116/58=2 mol C4H10 1.66*1024/6*1023=1.1 mol O2

2 mol C4H10 produces 10*2/2=10 mol H2O

1.1 mol O2 produces 10*1.1/13 =0.85 mol H2O

O2 limits Multiply down to calculate:grams H2O= 0.85*18=15.3 g

U-Try-It on your own Clicker Examples

2C4H10 + 13O2-------- 8CO2 + 10H2O Given 0.5 mol of C4H10 and 6.5 mol of O2, how many mol of CO2 will form?

8 4 2 1

25% 25%25%25%

2C4H10 + 13O2-------- 8CO2 + 10H2O 58 32 44 18 g/mol Given 10 g of C4H10 and 100 g of O2, how many mol of H2O will form?

A. 2.4

B. 0.17

C. 0.86

D. 3.12

E.0.0342.4

0.170.86

3.120.034

20% 20% 20%20%20%

2C4H10 + 13O2-------- 8CO2 + 10H2O 58 32 44 18 g/mol Given 2*1022 molecules of C4H10 and 50 g of O2, how many g of CO2 will form?

1 mol count =6*1023 A. 5.8 g

B. 0.133 g

C. 4.22 g

D. 1.2 g

0.133 g

4.22 g 1.2 g

25% 25%25%25%

% composition problems and combustion analysis (pp. 94-103) Reaction balancing (pp. 105-108)...

Documents