Transcript Ch4.Stoichiometry

Chemical
Reactions
Chapter 4
Stoichiometry
1
Balancing Chemical Equations
“Matter is conserved in
chemical change”
Antoine Lavoisier, 1789
An equation must be balanced:
It must have the same number
of atoms of each kind on both sides
4
Combustion Reactions

In combustion, a hydrocarbon or C–H–O fuel
combines with O2 to form CO2 and H2O
__ CH4 + __ O2  __ CO2 + __ H2O
1 CH4 + 2 O2  1 CO2 + 2 H2O
Balanced equation shows 1 C, 4 H, and 4 O on each side

If N or S are in the formula for the fuel, assume it
is oxidized to NO2 or SO2
6
Example 4-2
Write a balanced equation for the complete
combustion of glycerol, C3H8O3
 Write a balanced equation for the complete
combustion of thiosalicylic acid, C7H6O2S

8
Stoichiometry
Stoichiometry is chemical accounting
 The heart of stoichiometry is the mole ratio given
by the coefficients of the balanced equation

9
Stoichiometry
Stoichiometry is chemical accounting
 The heart of stoichiometry is the mole ratio given
by the coefficients of the balanced equation

moles A
mole ratio
moles B
moles A
moles B
10
Example 4-3B

How many moles of Ag are produced in the
decomposition of 1.00 kg of silver (I) oxide:
2 Ag2O (s)  4 Ag (s) + O2 (g)
12
Example 4-6B
The model problem describes
an Al-Cu alloy composed of
93.7% Al and 6.3% Cu by
mass, with a density of 2.85
g/cm3. The Al (but not the Cu)
reacts with HCl:
2 Al (s) + 6 HCl (aq)  2 AlCl3 (aq) + 3 H2 (g)
How
many grams of Cu are present in a sample of
alloy that yields 1.31 g H2 when it reacts with HCl?
18
Example 4-7B

A vinegar contains 4.0% HC2H3O2 by mass and
has a density of 1.01 g/mL. It reacts with sodium
hydrogen carbonate:
HC2H3O2 (aq) + NaHCO3 (s) 
NaC2H3O2 (aq) + H2O (l) +
CO2 (g)
How many grams of CO2 are produced by the
reaction of 5.00 mL of this vinegar with NaHCO3?
20
Chemical Reactions in Solution

Most reactions occur in aqueous solution
SOLUTE is the substance to be dissolved in solution
 SOLVENT is the substance (often a liquid) the solute
dissolves in


The concentration of the solution is
Molarity (M) = moles solute
L solution
21
Example 4-8B

15.0 mL of concentrated acetic acid, HC2H3O2
(d = 1.048 g/mL), are dissolved in enough water to
produce 500.0 mL of solution. What is the
concentration of the solution?
23
Example 4-9B

How many grams of Na2SO4 • 10 H2O are needed
to prepare 355 mL of 0.445 M Na2SO4?
25
Dilution problems
It is common to prepare a solution by diluting a
more concentrated solution (the stock solution).
 The moles of solute taken from the stock solution
are given by moles solute = volume x molarity
 All the solute taken from the stock appears in the
diluted solution, so moles solute are constant:
VstockMstock = VdiluteMdilute

26
Example 4-10A

15.00 mL of 0.450 M K2CrO4 solution are diluted
to 100.00 mL. What is the concentration of the
dilute solution?
27
Example 4-10B

After being left out in an open beaker, 275 mL of
0.105 M NaCl has evaporated to only 237 mL.
What is the concentration of the solution after
evaporation?
28
Stoichiometry in Solution

Stoichiometry in solution is just the same as for
mass problems, except the conversion into or out
of moles uses molarity instead of molar mass:
grams A
grams B
moles A
mL A
mole ratio
moles B
moles A
moles B
mL B
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Example 4-11B
K2CrO4 (aq) + 2 AgNO3 (aq)  Ag2CrO4 (s) + 2 KNO3 (aq)

How many mL of 0.150 M AgNO3 must react with
excess K2CrO4 to produce exactly 1.00 g
Ag2CrO4?
31
Limiting reactant
In a given reaction, often there is not enough of
one reactant to use up the other reactant completely
 The reactant in short supply LIMITS the quantity
of product that can be formed

32
33
Goldilocks Chemistry

Imagine reacting different amounts of Zn with
0.100 mol HCl:
Zn (s) + 2 HCl (aq)  ZnCl2 (aq) + H2 (g)
Mass Zn
Moles Zn
Moles HCl
Ratio mol HCl
mol Zn
Rxn 1
6.54 g
0.100 mol
0.100 mol
1.00
Rxn 2
3.27 g
0.0500 mol
0.100 mol
2.00
Rxn 3
1.31 g
0.0200 mol
0.100 mol
5.00
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Limiting reactant problems

The easiest way to do these is to do two
stoichiometry calculations

Find the amount of product possible from each reactant
The smaller answer is the amount of product you
can actually make (you just ran out of one reactant)
 The reactant on which that answer was based is the
limiting reactant

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Example 4-13A

When 215 g P4 react with 725 g Cl2
P4 (s) + 6 Cl2 (g)  4 PCl3 (l) (example 4-12A)
which reactant is in excess and what mass of that
reactant remains after the reaction is finished?
38
Example 4-13B

12.2 g H2 and 154 g O2 are allowed to react.
Identify the limiting reactant, which gas remains
after the reaction, and what mass of it is left over.
2 H2 (g) + O2 (g)  2 H2O (l)
39
Percent Yield

In real experiments we often do not get the amount
of product we calculate we should, because
the reactants may participate in other reactions (side
reactions) that produce other products (by-products)
 The reaction often does not go to completion.


Percent yield tells the ratio of actual to theoretical
amount formed.
40
Percent Yield
Suppose you calculate that a reaction will produce
50.0 g of product. This is the theoretical yield.
 The reaction actually produces only 45.0 g of
product . This is the actual yield.


Percent yield = 45.0 g (actual) x 100 = 90.0%
50.0 g (theoretical)
41
Example 4-14B

What is the percent yield if 25.0 g P4 reacts with
91.5 g Cl2 to produce 104 g PCl3:
P4 (s) + 6 Cl2 (g)  4 PCl3 (l)
43
Example 4-15B

What mass of C6H11OH should you start with to
produce 45.0 g C6H10 if the reaction has 86.2%
yield and the C6H11OH is 92.3% pure:
C6H11OH (l)  C6H10 + H2O (l)
45
Exercise 26

Balance these equations by inspection
(NH4)2Cr2O7 (s)  Cr2O3 (s) + N2 (g) + H2O (g)
 NO2 (g) + H2O (l)  HNO3 (aq) + NO (g)
 H2S (g) + SO2 (g)  S (g) + H2O (g)
 SO2Cl2 + HI  H2S + H2O + HCl + I2

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Exercise 30

Write balanced equations for these reactions:
Sulfur dioxide gas with oxygen gas to produce sulfur
trioxide gas
 Solid calcium carbonate with water and dissolved
carbon dioxide to produce aqueous calcium hydrogen
carbonate
 Ammonia gas and nitrogen monoxide gas to produce
nitrogen gas and water vapor

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Exercise 32
3 Fe (s) + 4 H2O (g)  Fe3O4 (s) + H2 (g)

How many moles of H2 can be produced from 42.7 g Fe
and excess steam?
 How many grams of H2O are consumed in the
conversion of 63.5 g Fe to Fe3O4?
 If 7.36 mol H2 are produced, how many grams of Fe3O4
must also be produced?

48
Exercise 36

Silver oxide decomposes above 300 °C to yield
metallic silver and oxygen gas. 3.13 g impure
silver oxide yields 0.187 g O2. Assuming there is
no other source of O2, what is the % Ag2O by mass
in the original sample?
49
Exercise 42

How many grams of CO2 are produced in the
complete combustion of 406 g of a bottled gas that
consists of 72.7% C3H8 (propane) and 27.3%
C4H10 (butane), by mass?
50
Exercise 45

What are the molarities of these solutes?
150.0 g sucrose (C12H22O11) in 250.0 mL aqueous
solution
 98.3 mg of 97.9% pure urea, CO(NH2)2, in 5.00 mL
aqueous solution
 12.5.0 mL methanol (CH3OH, density = 0.792 g/mL) in
15.0 L aqueous solution

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Exercise 52

After 25.0 mL of aqueous HCl solution is diluted
to 500.0 mL, the concentration of the diluted
solution is found to be 0.085 M HCl. What was
the concentration of the original HCl solution?
52
Exercise 56
 Ca(OH)2 (s) + 2 HCl (aq)  CaCl2 (aq) + 2 H2O (l)


How many grams of Ca(OH)2 will react completely with 415 mL
of 0.477 M HCl?
How many kilograms of Ca(OH)2 will react with 324 L of an
HCl solution that is 24.28% HCl by mass, density = 1.12 g/mL?
53
Exercise 63

0.3126 g oxalic acid, H2C2O4, is exactly
neutralized by 26.21 mL of a NaOH solution.
What is the concentration of the NaOH solution?
H2C2O4
+ 2 NaOH  Na2C2O4 + 2 H2O
54
Exercise 70

Chlorine can be generated by heating calcium
hypochlorite and hydrochloric acid to form
chlorine gas, calcium chloride, and water. If 50.0 g
Ca(OCl)2 and 275 mL 6.00 M HCl react, how
many grams of Cl2 gas form? Which reactant is
left over, and how much (in grams)?
55
Exercise 72
2 C6H5NO2 + 4 C6H14O4  (C6H5N)2 + 4 C6H12O4 + 4 H2O
nitrobenzene

triethylene
glycol
azobenzene
If 0.10 L nitrobenzene (d = 1.20 g/mL) react with
0.30 L triethylene glycol (d = 1.12 g/mL) to form
55 g azobenzene, find
Theoretical yield
 Actual yield
 Percent yield

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