Dr. S. M. Condren Chapter 4 Chemical Equations and Stoichiometry.

Dr. S. M. Condren

Chapter 4

Chemical Equations

and

Stoichiometry

Dr. S. M. Condren

CHEMICAL REACTIONS

Reactants: Zn + IReactants: Zn + I22 Product: Zn IProduct: Zn I22

Dr. S. M. Condren

Chemical EquationsDepict the kind of reactants and

products and their relative amounts in a reaction.

4 Al(s) + 3 O2(g) ---> 2 Al2O3(s)

The numbers in the front are called

stoichiometric coefficientsThe letters (s), (g), and (l) are the physical

states of compounds, (aq) refers to an aqueous or water solution.

Dr. S. M. Condren

The Mole and Chemical Reactions:

The Nano-Macro Connection

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

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

4 g H2 32 g O2 36 g H2O

Dr. S. M. Condren

Stoichiometry

stoi·chi·om·e·try noun1. Calculation of the quantities of reactants

and products in a chemical reaction.2. The quantitative relationship between

reactants and products in a chemical reaction.

Dr. S. M. Condren

4 Al(s) + 3 O2(g) ---> 2 Al2O3(s)

This equation means

4 Al atoms + 3 O2 molecules ---give--->

2 “molecules” of Al2O3

4 moles of Al + 3 moles of O2 ---give--->

2 moles of Al2O3

Chemical Equations

Dr. S. M. Condren

• Because the same atoms are present in a reaction at the beginning and at the end, the amount of matter in a system does not change.

• The Law of the Conservation of Matter

Demo of conservation of matter

2HgO(s) ---> 2 Hg(liq) + O2HgO(s) ---> 2 Hg(liq) + O22(g)(g)

Chemical Equations

Dr. S. M. Condren

Combination Reaction

PbNO3(aq) + K2CrO4(aq) PbCrO4(s) + 2 KNO3(aq)

Colorless yellow yellow colorless

Dr. S. M. Condren

Because of the principle of the conservation of matter,

an equation must be balanced.

It must have the same number of atoms of the same kind on both sides.

Chemical Equations

Lavoisier, 1788Lavoisier, 1788

Dr. S. M. Condren

___ Al(s) + ___ Br2(liq) ---> ___ Al2Br6(s)

BalancingEquations

Dr. S. M. Condren

____C3H8(g) + _____ O2(g) ---->

_____CO2(g) + _____ H2O(g)

____B4H10(g) + _____ O2(g) ---->

___ B2O3(g) + _____ H2O(g)

BalancingEquations

Dr. S. M. Condren

EXAMPLE How much H2O, in moles results from burning an excess of H2 in 3.3 moles of O2?

H2 + O2 -----> H2O

#mol H2O =

22

(3.3 mol O2)(2 mol H2O)

(1 mol O2)

Now we need the stoichiometric factor

= 6.6 mol H2O

Dr. S. M. Condren

Stoichiometric Roadmap

Dr. S. M. Condren

EXAMPLE How much H2O, in grams results from burning an excess of H2 in 3.3 moles of O2?

H2 + O2 -----> H2O

#g H2O =

22

(3.3 mol O2)(2 mol H2O)

(1 mol O2)

= 1.2x102 g H2O

(18.0 g H2O)

(1 mol H2O)

Dr. S. M. Condren

EXAMPLE How much H2O, in grams results from burning an excess of H2 in 3.3 grams of O2?

H2 + O2 -----> H2O#g H2O =

22

(1 mol O2) (2 mol H2O)

(1 mol O2)

= 3.7 g H2O

(18.0 g H2O)

(1 mol H2O)

(3.3 g O2)

(32.g O2)

Dr. S. M. Condren

Thermite Reaction

Dr. S. M. Condren

Thermite Reaction

Fe2O3(s) + 2Al(s) Al2O3(s) + 2 Fe(l)

Dr. S. M. Condren

Thermite Reaction

Dr. S. M. Condren

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

Dr. S. M. Condren

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/yard and that one inch of the rails will be covered by an additional 10% mass of iron.

The mass of iron in 1 inch of this rail is:#g/in = (132 #/yard) (1 yard/36 in) (454 g/#)

= 1.67 X 103 g/in

The mass of iron in a weld adding 10% mass:#g = (1.67 X 103 g) (0.10) = 167 g

Dr. S. M. Condren

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

The mass of iron in a weld adding 10% mass:#g = (1.67 X 103 g) (0.10) = 167 g

Balanced chemical equation:Fe2O3 + 2 Al ---> 2 Fe + Al2O3

What mass of Fe2O3 is required for the thermite process?

#g Fe2O3 = (167 g Fe) (1 mol Fe)---------------- (55.85 g Fe)

(1 mol Fe2O3)------------------- (2 mol Fe)

(159.7 g Fe2O3)---------------------- (1 mol Fe2O3)

= 238 g Fe2O3

Dr. S. M. Condren

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

The mass of iron in a weld adding 10% mass:#g = (1.67 X 103 g) (0.10) = 167 g

Balanced chemical equation:Fe2O3 + 2 Al ---> 2 Fe + Al2O3

What mass of Al is required for the thermite process?

#g Al = (167 g Fe) (1 mol Fe)---------------- (55.85 g Fe)

(2 mol Al)--------------(2 mol Fe)

(26.9815 g Al)------------------- (1 mol Al)

= 80.6 g Al

Dr. S. M. Condren

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/yard and that one inch of the rails will be covered by an additonal 10% mass of iron.

The mass of iron in a weld adding 10% mass:

#g Fe = 167 g Fe

#g Fe2O3 = 238 g Fe2O3

#g Al = 80.6 g Al

Dr. S. M. Condren

• In a given reaction, there is not In a given reaction, there is not enough of one reagent to use up enough of one reagent to use up the other reagent completely.the other reagent completely.

• The reagent in short supply The reagent in short supply

LIMITSLIMITS the quantity of product the quantity of product that can be formed.that can be formed.

Reactions Involving aLIMITING REACTANT

Dr. S. M. Condren

LIMITING REACTANTS

Reactantseactants ProductsProducts

2 NO(g) + O2 (g) 2 NO2(g)

Limiting reactant = ___________Limiting reactant = ___________Excess reactant = ____________Excess reactant = ____________

Dr. S. M. Condren

EXAMPLEWhat is the number of moles of Fe(OH)3 (S) that can be produced by allowing 1.0 mol Fe2S3, 2.0 mol H2O, and 3.0 mol O2 to react?

2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)

balanced equation relates:2Fe2S3(S) <=> 6H2O(l) <=> 3O2(g)

have only:1Fe2S3 (S) <=> 2H2O(l) <=> 3O2(g)

not enough H2O to use all Fe2S3

plenty of O2

Dr. S. M. Condren

EXAMPLE What is the number of moles of Fe(OH)3 (S) that can be produced by allowing 1.0 mol Fe2S3, 2.0 mol H2O, and 3.0 mol O2 to react?

2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)

if use all Fe2S3:

(1.0 mol Fe2S3) (4 mol Fe(OH)3) #mol Fe(OH)3 = ------------------------------------------

(2 mol Fe2S3)

= 2.0 mol Fe(OH)3

Dr. S. M. Condren

EXAMPLE: What is the number of moles of Fe(OH)3 (S) that can be produced by allowing 1.0 mol Fe2S3, 2.0 mol H2O, and 3.0 mol O2 to react?

2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)

if use all H2O:

(2.0 mol H2O) (4 mol Fe(OH)3) #mol Fe(OH)3 = -----------------------------------------

(6 mol H2O)= 1.3 mol Fe(OH)3

Dr. S. M. Condren

EXAMPLE: What is the number of moles of Fe(OH)3 (S) that can be produced by allowing 1.0 mol Fe2S3, 2.0 mol H2O, and 3.0 mol O2 to react?

2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)

if use all O2

(3.0 mol O2) (4 mol Fe(OH)3) #mol Fe(OH)3 = ---------------------------------------

(3 mol O2)

= 4.0 mol Fe(OH)3

Dr. S. M. Condren

EXAMPLE: What is the number of moles of Fe(OH)3 (S) that can be produced by allowing 1.0 mol Fe2S3, 2.0 mol H2O, and 3.0 mol O2 to react?

2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)

1.0 mol Fe2S3 => 2.0 mol Fe(OH)3

2.0 mol H2O => 1.3 mol Fe(OH)3

3.0 mol O2 => 4.0 mol Fe(OH)3

least amount

Since 2.0 mol H2O will produce only 1.3 mol Fe(OH)3,then H2O is the limiting reactant.

Thus the maximum number of moles of Fe(OH)3 that can be produced by this reaction is 1.3 moles.

Dr. S. M. Condren

Theoretical Yield

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

Dr. S. M. Condren

Actual Yield

amount of product actually produced in a reaction

Dr. S. M. Condren

Percent Yield

actual yield% yield = --------------------- * 100 theoretical yield

Dr. S. M. Condren

EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4?

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

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

Dr. S. M. Condren

EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4?

2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) to calculate the theoretical yield, use the net equation

for the overall process

(1.00 kg Cl2) #kg N2H4 = ---------------------

Dr. S. M. Condren

EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4?

2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) to calculate the theoretical yield, use the net equation

for the overall process

(1.00 kg Cl2) (1000 g Cl2)#kg N2H4 = -----------------------------------

(1 kg Cl2) metric conversion

Dr. S. M. Condren

EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4?

2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) to calculate the theoretical yield, use the net equation

for the overall process

(1.00) (1000 g Cl2) (1 mol Cl2) #kg N2H4 = -----------------------------------------

(1) (70.9 g Cl2) molar mass

Dr. S. M. Condren

EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4?

2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) to calculate the theoretical yield, use the net equation

for the overall process

(1.00)(1000)(1 mol Cl2) #kg N2H4 = -----------------------------------------

(1)(70.9)

Dr. S. M. Condren

EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4?

2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) to calculate the theoretical yield, use the net equation

for the overall process

(1.00)(1000)(1 mol Cl2)(1 mol N2H4) #kg N2H4 = -------------------------------------------------

(1) (70.9) (1 mol Cl2)

Dr. S. M. Condren

EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4?

2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) to calculate the theoretical yield, use the net equation

for the overall process

(1.00)(1000)(1)(1 mol N2H4) #kg N2H4 = -------------------------------------------------

(1) (70.9)(1)

Dr. S. M. Condren

EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4?

2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) to calculate the theoretical yield, use the net equation

for the overall process

(1.00)(1000)(1)(1 mol N2H4) (32.0 g N2H4) #kg N2H4 = --------------------------------------------------------

(1)(70.9) (1) (1 mol N2H4)

molar mass

Dr. S. M. Condren

EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4?

2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) to calculate the theoretical yield, use the net equation

for the overall process

(1.00)(1000)(1)(1) (32.0 g N2H4)(1 kg N2H4)#kg N2H4 = ----------------------------------------------------------

(1)(70.9)(1)(1) (1000 g N2H4)

metric conversion

Dr. S. M. Condren

EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4?2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) to calculate the theoretical yield, use the net equation

for the overall process(1.00)(1000)(1)(1) (32.0)(1 kg N2H4)

#kg N2H4 = ----------------------------------------------------------(1)(70.9)(1)(1)(1000)

Dr. S. M. Condren

EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4?2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) to calculate the theoretical yield, use the net equation

for the overall process(1.00)(1000)(1)(1) (32.0)(1 kg N2H4)

#kg N2H4 = ----------------------------------------------------------(1)(70.9)(1)(1)(1000)

= 0.451 kg N2H4

Dr. S. M. Condren

EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4?

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

(b) actual yield

(0.299 kg product) # kg N2H4 = --------------------------

Dr. S. M. Condren

EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4?2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) theoretical yield #kg N2H4 = 0.451 kg N2H4

(b) actual yield (0.299 kg product) (98.0 kg N2H4)

# kg N2H4 = -------------------------------------------- (100 kg product)

purity factor

Dr. S. M. Condren

EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4?2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) theoretical yield #kg N2H4 = 0.451 kg N2H4

(b) actual yield (0.299 kg product) (98.0 kg N2H4)

# kg N2H4 = -------------------------------------------- (100 kg product)

= 0.293 kg N2H4

Dr. S. M. Condren

EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4?

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

(b) actual yield # kg N2H4 = 0.293 kg N2H4

Dr. S. M. Condren

EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4?2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) theoretical yield #kg N2H4 = 0.451 kg N2H4

(b) actual yield # kg N2H4 = 0.293 kg N2H4

(c) percent yield 0.293 kg

% yield = -------------- X 100 = 65.0 % yield 0.451kg