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Free Group Study Tutoring • Monday 5-6pm • Tuesday 5-6pm • Friday 11am-12pm • Tutoring will begin Friday Sept. 16 th at 11am in the Learning Assistance Center! • For more information call the LAC at: 562-985-5350

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Free Group Study Tutoring

• Monday 5-6pm • Tuesday 5-6pm• Friday 11am-12pm• Tutoring will begin Friday Sept. 16th at

11am in the Learning Assistance Center!• For more information call the LAC at:

562-985-5350

Stoichiometry

stoi·chi·om·e·try nounLiterally – measuring the componentsCalculations of the quantitative relationships between reactants and products in a chemical reaction.

The Mole and Chemical Reactions:The Macro-Nano Connection

2 H2 + O2 2 H2O 2 H2 molecules 1 O2 molecule 2 H2O molecules

2 moles H2 molecules 1 mole O2 molecules 2 moles H2O molecules

4.0 g H2 32.0 g O2 36.0 g H2O

Stoichiometric Relationships

EXAMPLE How much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together?

Assume that the rail is 132 lb Fe/yard and that the weld is 1/10 inch wide.

weld

Photo by Mike Condren

The mass of iron in 1/10 inch of this rail is:

g166lb1

g454

yd1

lb132

in36

yd1in10

1

Fe2O3 + 2 Al Al2O3 +2 Fe

3232

3232 OFeg237OFemol1

OFeg68.159

Femol2

OFemol1

Feg84.55

Femol1Feg166

Alg.80Almol1

Alg98.26

Femol2

Almol2

Feg84.55

Femol1Feg166

1. C

2. Ge

3. Si

4. Sn

5. Ti

Limiting Reactant

reactant that limits the amount of product that can be produced

Limiting Reactant

EXAMPLE What is the number of moles of CaSO4 (S) that can be produced by allowing 1.0 mol SO2, 2.0 mol CaCO3, and 3.0 mol O2 to react?

2SO2(g) + 2CaCO3(s) + O2(g) 2CaSO4(S) + 2CO2(g)

balanced equation relates:2SO2(g) 2CaCO3(s) O2(g)

have only:1SO2(g) 2CaCO3(s) 3O2(g)

not enough SO2 to use all of the CaCO3 or the O2

not enough CaCO3 to use all of the O2

SO2 is the limiting reactant

Compare Amounts of Product

42

42 CaSOmol1

SOmol2

CaSOmol2SOmol1

43

43 CaSOmol2

CaCOmol2

CaSOmol2CaCOmol2

42

42 CaSOmol6

Omol1

CaSOmol2Omol3

SO2 limits the amount of product formed

Mass of Product

• 95.0 g of chlorine and 27.0 g of phosphorus react to form PCl3. What mass of PCl3 is formed?

P4 (s) + 6 Cl2 (g) 4 PCl3 (l)

33

3

2

3

2

22 PClg7.122

PClmol1

PClg32.137

Clmol6

PClmol4

Clg90.70

Clmol1Clg0.95

33

3

4

3

4

44 PClg7.119

PClmol1

PClg32.137

Pmol1

PClmol4

Pg88.123

Pmol1Pg0.27

27.0g phosphorus gives the smaller yield. Phosphorus is the limiting reactant it limits the amount of product formed

Excess Reactant

• We can also calculate the amount of the non-limiting reactant that is used up, and thus how much is left over

23

2

4

2

4

44 Clg72.92

PClmol1

Clg90.70

Pmol1

Clmol6

Pg88.123

Pmol1Pg0.27

92.72 g Cl2 are required to react with all the P4

Thus 95.0-92.72=2.3 g Cl2 are left over as excess

Theoretical Yield

The amount of product produced by a reaction based on the amount of the limiting reactant

Actual Yield

The amount of product actually produced in a reaction

Percent Yield

actual yield% yield = 100% theoretical yield

For example, if only 103.5 g of PCl3

were actually produced

yield%5.86%100PClg7.119

PClg5.103yield%

3

3

EXAMPLE A rocket fuel, hydrazine, is produced by a reaction of Cl2 with excess NaOH and NH3. (a) What theoretical yield can be produced from 1.00 kg of Cl2?

2NaOH + Cl2 + 2NH3 N2H4 + 2NaCl + 2H2O

(a) to calculate the theoretical yield, use the net equation for the overall process

(1.00 kg Cl2) (1 kmol Cl2)

(70.9 kg Cl2) molar mass

(1 kmol N2H4)

(1 kmol Cl2)balancedequation

(32.0 kg N2H4)

(1 kmol N2H4)molar mass

= 0.451 kg N2H4

EXAMPLE (b) What is the actual yield if 0.299 kg of 98.0% N2H4 is produced for every 1.00 kg of Cl2?

(98.0 kg N2H4)

(100 kg product)purity factor

(0.299 kg product) = 0.293 kg N2H4

2NaOH + Cl2 + 2NH3 N2H4 + 2NaCl + 2H2O(a) theoretical yield #kg N2H4 = 0.451 kg N2H4

(b) actual yield

EXAMPLE (c) What is the percent yield of pure N2H4?

2NaOH + Cl2 + 2NH3 N2H4 + 2NaCl + 2H2O(a) theoretical yield = 0.451 kg N2H4

(b) actual yield = 0.293 kg N2H4

(c) percent yield 0.293 kg

% yield = 100% = 65.0 % yield 0.451kg

1. 16 %

2. 32 %

3. 50 %

4. 65 %

Combustion Analysis

Example Benzoic acid is known to contain only C, H, and O. A 6.49-mg sample of benzoic acid was burned completely in a C-H analyzer. The increase in the mass of each absorption tube showed that 16.4-mg of CO2 and 2.85-mg of H2O formed. What is the empirical formula of benzoic acid?

(16.4-mg of CO2 )(12.01-mg C) # mg C = = 4.48-mg C (44.01-mg CO2)

(2.85-mg of H2O )(2.01-mg H) # mg H = = 0.318-mg H (18.01-mg H2O)

# mg O = 6.49 - 4.48 - 0.318 = 1.69 mg O

Empirical Formula

72Cmmol53.3106.0/Cmmol373.0Cmg01.12

Cmmol1Cmg48.4

62Hmmol99.2106./Hmmol315.0Hmg008.1

Hmmol1Hmg318.0

22Ommol00.1106./Ommol106.0Omg00.16

Ommol1Omg69.1

C7H6O2