Page 12

2K + Cl2  2KCl Type: Synthesis

2AlBr3 + 3 Na2(CO3) Al2(CO3)3 +6 NaBr Type: Double
replacement

2C2H6 + 7O2  4 CO2 +6 H2O Type:Combustion
3Cu2(CrO4) + 2 Fe  Fe2(CrO4)3 + 6Cu
Type: Single replacement


Mg(CO3) MgO + CO2 Type: decomposition

3H2 + Fe2S3  3 H2S + 2 Fe Type: Single replacement
 2AlBr3
(aq)+ 3 Na2(CO3) (aq) Al2(CO3)3 (s)
+6 NaBr(aq)
Precipitate!
(The insoluble
product)
 3H2
+ Fe2S3  3 H2S + 2 Fe will not occur.
Iron is more reactive than hydrogen. (Refer
to the activity series on the back of your
periodic table). The more reactive element
will be in the compound.
 Copper(II)
chloride reacts with iron to
produce iron(III) chloride and copper metal.
Skeleton: CuCl2 + Fe  FeCl3 +Cu
Balanced: 3CuCl2 + 2Fe  2FeCl3 +3Cu
Type of reaction : Single replacement
 Hydrogen
gas and bromine liquid react to
yield hydrogen bromide.
Skeleton: H2 + Br2  HBr
Balanced: H2 + Br2  2HBr
Type of reaction : Synthesis
 Carbon
tetrahydride reacts with oxygen to
produce carbon dioxide and water vapor.
Skeleton: CH4 + O2  CO2+ H2O
Balanced: CH4 + 2O2  CO2 + 2H2O
Type of reaction : Combustion
Pages 13 and 14
___HCl + ___Ca(OH)2  ___CaCl2 + ___H2O
Balance: 2HCl + 1Ca(OH) 2 1CaCl2 +2H2O
Use the mole ratio from the balanced equation to
convert from moles of HCl to moles of CaCl2 :
2HCl + 1Ca(OH) 2 1CaCl2 +2H2O
1.53 mol HCl
1 mol CaCl2
2 mol HCl
= .765 mol CaCl2
___HBr + ___Fe  ___FeBr3 + ___H2
Balance: 6HBr + 2Fe 2FeBr3 + 3H2
 Use the mole ratio from the balanced equation to
convert from moles of Fe to moles of FeBr3:
 Use molar mass to convert between moles & grams.
6HBr + 2Fe 2FeBr3 + 3H2
2.05 mol Fe 2 mole FeBr3 295.557g = 606g FeBr3
2 mole Fe
1 mole FeBr3
Molar mass of FeBr3
___Na + ___Cl2  ___NaCl
Balance: 2Na + 1Cl2  2NaCl
 Use
the mole ratio from the balanced equation
to convert from moles of Na to moles of Cl2:
 Use molar mass to convert between moles &
grams.
2Na + 1Cl2  2NaCl
2.3g Na
1 mol Na
2 mol NaCl
22.990g Na
2 mol Na
Molar mass of Na
58.44g NaCl
1 mol NaCl
Molar mass of NaCl
= 5.8g NaCl
A. If a sample containing 18.1 grams of NH3 reacted with 90.4
grams of copper(II) oxide, which is the limiting reactant?
2NH3 +3CuO  1N2 + 3Cu + 2H2O
18.1g NH3
1 mol NH3 1 mol N2
28.014 g N2 = 14.9 g N2
17.031g NH3 2 mol NH3 1 mol N2
90.4g CuO
1 mol CuO
79.545g CuO
1 mol N2
28.014 g N2 = 10.6 g N2
3 mol CuO 1 mol N2
A. CuO is the limiting reactant.
B. How many grams of N2 will theoretically be formed? 10.6g N2
1C7H6O3 + 1C4H6O3  1C9H8O4 + 1C2H4O2
200.0g C7H6O3
1 mol C7H6O3
1 mol C9H8O4
138.052g C7H6O3
1 mol C7H6O3
Answer = 260.9g theoretical
Actual
x 100 =
Theoretical
231
260.9
x100
= 88.5%
% yield
180.069g C9H8O4 =
1 mol C9H8O4
1C6H6 + 1Br2  1C6H5Br + 1HBr
30.0g C6H6
1 mol C6H6
1 molC6H5Br
78.054g C6H6 1 mol C6H6
65.0g Br2
1 mol Br2
159.808g Br2
1 mol C6H5Br
1 mol Br2
156.95g C6H5Br
= 60.32gC6H5Br
1 mol C6H5Br
156.95g C6H5Br
= 63.8 C6H5Br
1 mol C6H5Br
C6H6 was the limiting reactant, so only 60.32 g of C6H5Br would be
theoretically produced.
C6H6 was the limiting reactant, so only 60.32 g of C6H5Br
would be theoretically produced.
Actual
Theoretical
42.3
60.32
x 100 =
x100 = 65.82%
% yield
Pages 15,16,17, 18
Matter
Pure Substances
Elements
Compounds
Mixtures
Homogeneous
Mixtures
Heterogeneous
Mixtures
a. matter – anything that has mass or
takes up space
b. physical property – property of
matter that can be observed or
measured without changing the
substance
 Examples: color, density, mass,
volume
c. extensive physical property – depends on
amount of substance present
 Examples: mass, volume, length
d. intensive physical property – does not
depends on the amount of a substance
 Examples: density, color, melting point,
boiling point
e. Chemical property- used to describe ability
Examples: reactivity, flammability, separating
mixtures
f. Physical change- alters the appearance but
does not change the composition of the
substance.
Examples: phase change, separating
mixtures, dissolving, evaporating
 a.
Peas and carrots- chromatography to
separate by color.
 b. charcoal powder and iron powderchromatography to separate by color.
 c. salt water- salt dissolves in water so you
would use crystallization
 d. pigments in green food coloring- if the
pigments are dissolved use crystallization, if
not dissolved use filtration
 e. rubbing alcohol and water- distillation to
separate by boiling points
 57.
Define chemical change – change occurs
when one or more substances undergoes a
chemical reaction to form a new substance
Examples: cooking, combustion, oxidation,
fizzes
 58.
List AND EXPLAIN the four indicators of a
chemical change.
a.Color change
c. gas evolution
b. formation of a precipitate d. odor
 a.
A student pours hydrochloric acid into a
test tube containing a white, crystalline
powder. The mixture begins to bubble and
the test tube begins to feel cold.
ENDOTHERMIC
 b.
A student pours HCl at 25.2 C into a test
tube containing a small metal strip. The
mixture begins to fizz and the temperature
of the mixture rises to 38.6 C.
EXOTHERMIC
Solids
Liquids
Gases
Compressibility
Incompressible
Incompressible
Compressible
Structure
Tightly packed
Loosely packed
particles
No attraction
between
particle
Motion
Particles
vibrate
Ability to flow
Move freely
ability to flow
Shape
Fixed shape
Takes shape of
container
Spread out in
container
Volume
Fixed shape
Fixed volume
Depends on
size of
container
A. melting- solid to liquid
B. Freezing- liquid to solid
C. Vaporization- liquid to gas
D. condensation- gas to liquid
E. Sublimation- solid to gas
F. deposition- gas to solid
critical pointanything
above this
point will be
a gas
Liquid
Solid
Gas
Triple point- point
where all 3 phases
coexist
 a.
Pure substance – uniform unchanging
composition, ex: elements and
compounds
 b.
Element- single type of atom ex: gold
(Au), Hydrogen (H)
 c.
Compound- more than one element
combined ex: NaCl
 d.
Mixture- combination of two or more
pure substances ex: salt water
 e.
Homogeneous mixture- constant
composition throughout and are always in
one phase
 f.
Solution- homogeneous mixture
 g.
Heterogeneous mixture- mixtures do not
blend together smoothly and the individual
substances remain distinct ex: colloid,
suspension
 h.
Colloid- one substance is suspended evenly
throughout another substance ex: milk, fog,
jello
 i.
Suspension-large substance particles are
suspended in another substance.
Ex: muddy water, paint
64. 85 grams
65. 130 grams
66. 50 oC
Pages 19,20
 67.
unsaturated solution
 68.
30g will dissolve, 20 grams will
remain at the bottom of the beaker
 69.
25 more grams will dissolve

Molarity = moles of solute
Liters of solution
Molarity units = M (concentration)
You may have to covert mL to L
Given mL
1L
1000 mL
You may have to covert grams to moles
in order to solve for Molarity
Given grams 1 mole
molar mass
 M1V1=
M2V2
M= Molarity
 V= Volume

Molarity = moles of solute
Liters of solution
 Convert Grams  moles
7.20g
1 mole = .087 moles
83 g
 Convert mL L
500mL
1L
= .500L
1000 mL
 Plug into equation:
Molarity = .087 = .18M
.500

Molarity = moles of solute
Liters of solution
Plug into equation:
2.5
= X = 3.375 mol
1.35

 Formula:
M1V1=M2V2
M = Molarity, V = volume
 Solving for V1
(5.0M)(V1)= (.25M)(100mL)

(5.0M)(V1)= 25
(V1)= 5mL
Multiply .25 x 100
Divide both sides by 5.0 to get (V1) by itself
 Formula:
M1V1=M2V2
M = Molarity, V = volume
 Solving for M2
(3.5M)(20mL)= (M2)(100mL)

70 = (M2)(100mL)
.7M= (M2)
Multiply 3.5 x 20
Divide both sides by 100 to get (M2) by itself