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Chemistry
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Name(s): __________________
CHM 1020 W5
Week 5 Assignment
Objective
In today’s lab, we will demonstrate a few different types of chemical reactions and be able to
observe whether a chemical reaction has taken place.
We will also learn how to use chemical equations
to describe chemical reactions.
Background/Theory
The Periodic Table shows over 100 elements. The chemical literature describes millions of compounds
that are known—some isolated from natural sources, some synthesized by laboratory workers. The
combination of chemicals, in the natural environment or the laboratory setting, involves chemical
reactions. The change in the way that matter is composed is a
chemical reaction
, a process wherein
reactants (or starting materials) are converted into products. The new products often have properties
and characteristics that are entirely different from those of the starting materials.
Four ways in which chemical reactions may be classified are combination, decomposition, single
replacement (substitution), and double replacement (metathesis).
Two elements reacting to form a compound is a
combination reaction
. This process may be described by
the general formula:
A
+
B
⟶
AB
The rusting of iron or the combination of iron and sulfur are good examples.
4Fe(s)
+
3O
2
(g)
⟶
2Fe
2
O
3
(s) (rust)
Fe(s)
+
S(s)
⟶
FeS(s)
Two compounds reacting together as in the example below also is a combination reaction.
CaO(s)
+
CO
2
(g)
⟶
CaCO
3
(s)
A compound which breaks down into elements or simpler components typifies the
decomposition
reaction
. This reaction has the general formula:
AB
⟶
A
+
B
Some examples of this type of reaction are the electrolysis of water into hydrogen and oxygen:
2H
2
O(l)
⟶
2H
2
(g)
+
O
2
(g)
and the decomposition of potassium iodate into potassium iodide and oxygen:
2KIO
3
(s)
⟶
2KI(s)
+
3O
2
(g)
The replacement of one component in a compound by another describes the
single replacement
(or
substitution
) reaction. This reaction has the general formula:
AB
+
C
⟶
CB
+
A
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CHM 1020 W5
Processes which involve oxidation (the loss of electrons or the gain of relative positive charge) and
reduction (the gain of electrons or the loss of relative positive charge) are typical of these reactions. Use
of Table 1, the activity series of common metals, enables
chemists to predict which oxidation-reduction
reactions are possible. A more active metal, one higher in the table, is able to displace a less active
metal, one listed lower in the table, from its aqueous salt. Thus aluminum metal displaces copper metal
from an aqueous solution of copper(II) chloride; but copper metal will not displace aluminum from an
aqueous solution of aluminum(III) chloride.
2Al(s)
+
3CuCl
2
(aq)
⟶
3Cu(s)
+
2AlCl
3
(aq)
Cu(s)
+
AlCl
3
(aq)
⟶
No Reaction
(
Note that Al is oxidized to Al
3+
and Cu
2+
is reduced to Cu.
)
Hydrogen may be displaced from water by a very active metal. Alkali metals are particularly reactive with
water, and the reaction of sodium with water often is exothermic enough to ignite the hydrogen gas
released.
2Na(s)
+
2HOH(l)
⟶
2NaOH(aq)
+
H
2
(g)
+
heat
(Note that Na is oxidized to Na
+
and H
+
is reduced to H
2
.)
Active metals, those above hydrogen in the series, are capable of displacing hydrogen from aqueous
mineral acids such as HCl or H
2
SO
4
; however, metals below hydrogen will not replace hydrogen. Thus
zinc reacts with aqueous solutions of HCl and H
2
SO
4
to release hydrogen gas, but copper will not.
Zn(s)
+
2HCl(aq)
⟶
ZnCl
2
(aq)
+
H
2
(g) Cu(s)
+
H
2
SO
4
(aq)
⟶
No reaction
Two compounds reacting with each other to form two different compounds describes double
replacement reactions.
This process has the general formula:
AB
+
CD
⟶
AD
+
CB
There are two replacements in the sense that A replaces C in CD and C replaces A in AB. This type of
reaction generally involves ions which form in solution either from the dissociation of ionic compounds
or the ionization of molecular compounds. The reaction of an aqueous solution of silver nitrate with an
aqueous solution of sodium chloride is a good
example. The products are sodium nitrate and silver
chloride. We know a reaction has taken place since the insoluble precipitate silver chloride forms and
separates from solution.
AgNO
3
(aq)
+
NaCl(aq)
⟶
NaNO
3
(aq)
+
AgCl(s) (White precipitate)
In general, a double replacement results if one combination of ions leads to a precipitate, a gas or an un-
ionized or very slightly ionized species such as water. In all of these reaction classes, it is very often
possible to use your physical senses to observe whether a chemical reaction has occurred. The
qualitative criteria may involve the formation of a gaseous product, the formation of a precipitate, a
change in color, or a transfer of energy.
In the next section, you will practice identifying which of the four classes or classification of chemical
reactions mentioned above does each chemical reaction represent.
Note, in addition to combination,
decomposition, single replacement and double replacement reactions, we also have combustion
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CHM 1020 W5
reactions, which you will explore in later assignments.
For now, each reaction listed will fall into one of
the four categories described here.
Practice determining reaction classification
For each of the reactions below, classify as a combination, decomposition, single replacement, or double
replacement.
1.
Ca(s)
+
Cl
2
(g)
⟶
CaCl
2
(s)
combination
2.
2Cu(s)
+
O
2
(g)
⟶
2CuO(s) combination
3.
Ca(NO
3
)
2
(aq)
+
H
2
SO
4
(aq)
⟶
2HNO
3
(aq)
+
CaSO
4
(s) Double
replacement
4.
NH
3
(aq)
+
HCl(aq)
⟶
NH
4
Cl(aq)
combination
5.
Hg(NO
3
)
2
(aq)
+
2NaI(aq)
⟶
HgI
2
(s)
+
2NaNO
3
(aq)
double replacement
6.
AgNO
3
(aq)
+
NaCl(aq)
⟶
AgCl(s)
+
NaNO
3
(aq)
Double replacement
7.
Zn(s)
+
H
2
SO
4
(aq)
⟶
ZnSO
4
(aq)
+
H
2
(g)
double
replacement
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CHM 1020 W5
8.
H
2
CO
3
(aq)
⟶
CO
2
(g)
+
H
2
O(l)
single replacement
9.
2H
2
O(l)
⟶
2H
2
(g)
+
2O
2
(g)
decomposition
10.
2Li(s)
+
2H
2
O(l)
⟶
2LiOH(aq)
+
H
2
(g)
double replacement
Procedure for Experiments
Combination/Synthesis Reactions
Watch the following Video
:
https://www.youtube.com/watch?v=w2ydd9rJHws
1.
For the reaction of magnesium and atmospheric (diatomic) oxygen write out the balanced
chemical equation:
2 Mg(s) + O
2
(g)
2 MgO(s)
2.
Next describe what magnesium metal looks like, and what the product identified above looks
like.
For example, was the magnesium reactant shiny or dull, was it malleable or brittle?
Magnesium reactant is silvery-white in colour and very light in weight. It's shiny malleable, and ductile.
While product Magnesium oxide appears as white solid, hygroscopic in nature
Decomposition Reaction
Watch the following video:
https://www.youtube.com/watch?v=_Y1alDuXm6A
1.
Write out the balanced chemical equation for the thermal decomposition of Mercury oxide:
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Name(s): __________________
CHM 1020 W5
2
HgO (s) -------->
2
Hg (s) + O
2
2.
What did the Mercury (II) oxide reactant look like, and what did the products of the
decomposition reaction look like?
Reactant:
The reactant was Mercury (II) oxide, and it looked like a red liquid. It was a solid at room temperature,
but once heated up it became a gas.
Products:
Mercury is silvery-white in color, has a metallic luster like that of silver or gold, and is a very heavy metal.
It is also highly toxic if ingested or inhaled. The other product of this reaction was oxygen gas; this gas is
colorless and odorless at room temperature but becomes a bright blue when exposed to light or flame!
3.
How do we know the gas released was oxygen?
The sample gas can be isolated and then sparked to induce ignition. Since O
2
can undergo combustion,
ignition should be observed if it is present.
Single Displacement Reaction
Watch the following video:
https://www.youtube.com/watch?v=k8UtR7akNec
1.
Write out the balanced chemical equation for single replacement reaction of silver nitrate and
copper:
2AgNO3(aq) + Cu(s) → 2Ag(s) + Cu(NO3)2(aq)
2.
Describe your observation, and clearly identify the appearances of reactants and products.
There is a color change. The colorless solution of silver nitrate changes to a blue solution. This due to the
reduction of copper to copper (II) ions which are blue in color. Copper displace silver from silver nitrate
to form copper nitrate and silver solid. The presence of copper (II) ions in copper nitrate solution
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CHM 1020 W5
solution is responsible for the blue color of the solution. In addition, there is formation of a white solid
(which is silver). Silver ions in silver nitrate are oxidized to silver metal which is a white solid.
Double Displacement Reaction
Watch the following video:
https://www.youtube.com/watch?v=X2mB-q2NQXY
1.
Write out the balanced chemical equation for the double replacement reaction of lead nitrate
and potassium iodide:
Pb(NO
3
)
2
(aq)
+
2
KI
(aq)
→
2
KNO
3
(aq)
+ PbI
2
(s)
2.
Describe the color of each aqueous solution of reactants lead nitrate and potassium iodide:
Lead nitrate is yellow
Potassium Iodine is white
3.
Describe the color and state of each product (note, there are two products you will need to
describe the color and state of).
Recall, by “state” we mean aqueous (aq), gas (g), liquid (l), and
solid (s).
The color of PbI
2
is
yellow.
Its state is
solid.
The solid state of KNO
3
is
white.
Its state is
solid.
1.
Balance the following chemical equation:
_2__FeBr
3
+ _3___H
2
SO
4
_____Fe
2
(SO
4
)
3
+ __6__HBr
2.
Balance the following chemical equation:
___C
7
H
16
+ __11__O
2
__7___CO
2
+ ___8___H
2
O
3. What is the formula weight (in amu) for ammonia, NH
3
?
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CHM 1020 W5
Molecular weight of Nitrogen is 14.01 g/ mol
Molecular weight of Hydrogen is 1.01 g/mol
NH3 has 1 N and 3 H
Therefore the total molecular weight is
14.01 + ( 3*1.01) = 17.04 g/ mol
17.04
amu
4.
How many molecules are present in one mole of water, H
2
O?
Express your answer in scientific
notation.
6.022 x 10^23
5. A person drinks 1.50 × 10
3
g of water, H
2
O, per day. How many moles is this?
It is given that
Mass of water given = 1.50x10
3
g
Also, we know that
Molar mass of water = 18g/mol
Number of moles = given mass/molar mass
Number of moles = 1.50x10
3
/18
15000/18 =
83.3 moles
6.
How many grams of CH
3
OH are in 0.42 mol CH
3
OH?
Molar mass of CH3OH= 32 g/mol
therefore, 1 mol CH3OH= 32 g of CH3OH
=> 0.42 mol CH3OH= 0.42 * 32 = 13.44 g of CH3OH
The mass is 13.44g of
CH3
OH
7.
The balanced equation
N
2
(
g
)
+
3
H
2
(
g
)
⟶
2
N H
3
(
g
)
means which of the following?
a.
One gram of nitrogen reacts with three grams of hydrogen to form two grams of ammonia
b.
One mole of nitrogen reacts with three moles of hydrogen to form two moles of ammonia
c.
The equation means both a and b
d.
The equation means neither a nor b
8.
Most of the industrial ammonia today is produced via the Haber Process.
The Chemical reaction is
one we’ve seen many times now:
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Name(s): __________________
CHM 1020 W5
N
2
(
g
)
+
3
H
2
(
g
)
⟶
2
N H
3
(
g
)
If 20 moles of ammonia (
N H
3
) are needed for a particular process, how many moles of diatomic
Nitrogen (
N
2
) are needed?
20 moles NH3 × (1 mole N2 / 2 moles NH3)
=
10 moles N2
Practice with Stoichiometry Calculations – Problem Solving Workflow
:
8.
How many grams of H
2
are required to produce 7.50 grams of ammonia NH
3
?
Use the balanced
equation (this time the equation is already balanced):
N
2
(
g
)
+
3
H
2
(
g
)
⟶
2
N H
3
(
g
)
N
2
+ 3H
2
--------> 2NH
3
molar mass of NH
3
= 17.031 g mol
-1
mass of 2 mol of NH
3
= 17.031 g mol
-1
x 2 mol
= 34.062 g
molar mass of H
2
= 2.015 g mol
-1
mass of 3 mol of H
2
= 2.015 g mol
-1
x 3 mol
= 6.045 g
To produce 34.062 g of NH
3
, 6.045 g of H
2
is used.
To produce 7.50 g of NH
3
, the H2 needed is:
= (6.045 g/34.062 g) x 7.50 g
= 0.17747 x 7.50 g
=1.33 g of H
2
is needed
.
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CHM 1020 W5
9.
If 54.7 grams of propane (C
3
H
8
) and 89.6 grams of oxygen (O
2
) are available in the balanced
combustion reaction below:
C
3
H
8
+
5
O
2
⟶
3
CO
2
+
4
H
2
O
a) Which reactant is limiting (show your work)?
O2
C
3
H
8
+ 5O
2
=====> 3CO
2
+ 4H
2
O
Moles of C
3
H
8
mass given = 54.7 g
molar mass of propane = 44.1 g/mol
moles of propane = mass (g)/molar mass = 54.7 g/44.1 g/mol = 1.24 mol
moles of Oxygen (O
2
)
mass given = 89.6 g
molar mass of O
2
= 32.0 g/mol
moles of O
2
= mass (g)/molar mass = 89.6 g/32.0 g/mol = 2.80 mol
Moles Of CO2 from Propane
moles of CO2 formed = 1.24 mol C3H8 (3 mol CO2/1 mol C3H8) =
3.72 mol CO2
Moles of CO2 from O2
moles of O2 = 2.80 mol
moles of CO2 = 2.80 mol O2 (3 mol CO2/5 mol O2) =
1.68 mol CO2
b)
What is the theoretical yield of
C O
2
in grams?
moles of CO2 = 1.68 mol
molar mass of CO2 = 44.01 g/mol
mass of CO2 produced = moles x molar mass = 1.68 mol x 44.01 g/mol =
73.9 g
theoretical yield of CO2 is 73.9 g.
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