Chemistry Laboratory Report 1 | Trường Đại học Quốc tế, Đại học Quốc gia Thành phố HCM

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Môn: Chemistry Laboratory (CH012IU)

Trường: Trường Đại học Quốc tế, Đại học Quốc gia Thành phố Hồ Chí Minh

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Tác giả:

Profile Mai Nguyệt
CHEMISTRY LABORATORY REPORT
EXPERIMENT 1: CHEMICAL REACTIONS
Group: Section: 4 1910
Group members: Lê Nguyễn Vân Anh
Hoàng Nguyễn Minh Châu
Lê Thanh Như Quỳnh
Trần Băng Châu
Tạ Thị Minh Thu
Date: February 18th, 2019.
Score:
International University, Vietnam National University - HCMC
GENERAL CHEMISTRY LABORATORY
2
I.Introduction:
Chemical changes result in the formation of new substances. When a chemical reaction
occurs, substances called reactants are transformed into different substances called products that
often have different appearances and different properties. Observable signs of chemical reactions
can be a change in color, the formation of a solid, the release of gas, and the production of heat and
light. We also learn how to classify chemical reactions. One classification system involves five
general types of reactions: synthesis, decomposition, single displacement, double displacement, and
combustion. Chemical reaction always happens in the living system, typically the formation and
transformation of matter.
II.Learning Objective:
-To perform different types of chemical reactions (acid-base, precipitate, gas forming,
complex compound forming, and oxidation-reduction reactions).
-To identify some of the products in these reactions and describe the chemical changes.
-To write and balance the chemical equations for the reactions observed.
III.Materials and Equipment:
Equipment
Test tubes
Test tube rack
Test tube holders
Beakers
Alcohol lamp
Looped wire
Distilled water bottle
Materials
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3
0.5M CuSO
4
2M NaOH
6M NaOH
0.1M AgNO
3
0.5M KBr
0.5M KI
0.1M KI
2M NH OH
4
3% H
2
O
2
2M H
2
SO
4
MnO
2
Saturated FeSO
4
0.1M NaNO
3
0.1M NaNO
2
0.1M KMnO4
C
2
H
5
OH
2M H
2
SO
4
Concentrated CH COOH
3
0.5M Na
2
SO
3
0.1M KSCN
2M KOH
0.5M K ]
4
[Fe(CN)
6
0.5M FeSO
4
0.5M FeCl
3
0.5M Al
2
(SO
4
)
3
2M HCl
0.5M LiCl
0.5M NaCl
0.5M KCl
0.5M CaCl
2
0.5M BaCl
2
0.5M K
2
Cr
2
O
7
96% H
2
SO
4
IV. Experimental Procedure: Data and Observations:
1. Reactions of Cu
2+
-First, put 10 drops of 0.5M CuSO into each of three test tubes.
4
-Then added 10 drops of 2M NaOH into the first test tube; 2M NH
4
OH into the second test tube;
0.5M K
4
[Fe(CN)
6
] into the third one.
- A few seconds later, added more 10 drops of 2M NaOH into the first test tube; 2M NH
4
OH into
the second test tube.
Reaction
Observation
Image
0.5M CuSO
4
+ 2M NaOH
-The blue precipitate
and colloidal
phenomena were
formed
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GENERAL CHEMISTRY LABORATORY
4
Explanation:
* CuSO is dissociated completely:
4
CuSO Cu + SO
4
→
2+
4
2-
* NaOH is also dissociated completely to form Hydroxide ion OH :
-
NaOH Na→
+
+ OH
-
*Ion Cu and OH in the solution combine together to form Cu(OH) , a blue precipitate.
2+
-
2
0.5M CuSO
4
+ 2M NH
4
OH
-First, the blue
precipitate dissolved
after being formed.
Then, the solution
turned into dark blue.
Explanation:
* CuSO is dissociated completely:
4
CuSO
4
→ Cu + SO
2+
4
2-
.
* NH OH is also dissociated completely:
4
NH
4
OH + OH → NH
4+
-
*Ion Cu
2+
and OH
-
in the solution combine together to form blue precipitate Cu(OH) because it has
2
very low solubility.
*Then, the remaining NH OH reacts with the precipitate Cu(OH) to form a
4
2
complex-compound [Cu(NH
3
)
4
]SO4 to form the dark blue of the solution.
0.5M CuSO
4
0.5M K
4
[Fe(CN)
6
]
-The brown
precipitate and
colloidal phenomena
were formed.
Explanation:
* Copper (II) sulfate CuSO is dissociated completely:
4
CuSO + SO4
4
→ Cu
2+
2-
Comments:
● The feature of Cu is it can react with OH form blue precipitation.
2+
-
● The solution Cu(OH)
2
also react with NH
4
OH to form a complex compound so reaction 2 has
the dark blue at the end.
● The last precipitate solution has brown color because the solution is compound consist of Fe
2+
● Double displacement reaction occurs when part of one reactant is replaced by part of another
reactant.
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5
● Comparing:
- The color of second precipitate is less dark than in practice.
- The color of third precipitate is just only brown unlike red-brown in practical.
2. Reactions of silver halides
-Prepared 0.5 M solutions of KCl, KBr, and KI.
-Then added 10 drops of 0.1M silver nitrate to 10 drops of each salt solution: KCl, KBr, and KI.
Observing the result
-Next, we divided each of the solutions equally into two test tubes labeled # 1-3 A and # 1-3 B
*With test tube #1-3 A: did not add anything, just observe.
*With test tube # 1-3 B: added 5 drops of 2M NH OH to each of test tubes.
4
-Waited till the end of the solution and recorded the observation.
Reaction
Observation
Chemical Equation
Image
0.5M KCl
+ 0.1M AgNO
3
-The white precipitate
was formed.
AgNO
3
+ KCl → AgCl + KNO
3
Explanation:
*KCl dissociate completely to form Cl
-
*In the solution, ion Cl
-
combine with ion Ag
+
to form AgCl which is a white precipitate because they
have low solubility.
0.5M KCl
+ 0.1M AgNO
3
+ 2M NH
4
OH
-The white precipitate
was formed.
-After adding NH
4
OH,
the precipitate was
dissolved slowly.
AgNO
3
+ KCl → AgCl + KNO
3
AgCl + 2NH
4
OH →
[Ag(NH
3
)
2
]Cl
+ H
2
O
Explanation:
*The precipitate AgCl reacts with Ammonium hydroxide NH OH to create the complex – compound
4
[Ag(NH
3
)
2
]Cl, which solute completely in the solution.
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6
0.5M KBr
+ 0.1M AgNO
3
-The light yellow
precipitate was
formed.
AgNO
3
+ KBr →AgBr + KNO
3
Explanation:
*KBr dissociate completely to form BRr
-
*In the solution, ion Br combine with ion Ag
-
+
to form AgBr which is a white-yellow precipitate because they have low solubility.
0.5M KBr
+ 0.1M AgNO
3
+ 2M NH
4
OH
-The light yellow
precipitate was
formed.
-After adding NH OH,
4
the precipitate was
dissolved a little.
AgNO
3
+KBr → AgBr + KNO
3
AgBr + 2NH OH →
4
[Ag(NH
3
)
2
]Br + H
2
O
Explanation:
*The precipitate AgBr reacts with NH OH to create the complex-compound [Ag(NH ]Br, which
4
3
)
2
solute completely in the solution.
0.5M KI
+ 0.1M AgNO
3
-The milky- white
yellow.
AgNO
3
+ KI → AgI + KNO
3
Explanation:
*KI dissociate completely to form I
-
*In the solution, ion I combine with ion Ag
-
+
to form AgI which is a light yellow precipitate because they have low solubility.
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7
0.5M KI
+ 0.1M AgNO3
+ 2M NH
4
OH
-The milky- white
yellow precipitate
appeared.
-After adding NH OH,
4
the precipitate
dissolved hardly.
AgI+2NH
4
OH → [Ag(NH
3
)
2
]I
+ H
2
O
Explanation:
*The precipitate AgI reacts with NH OH to create a milky yellow complex – compound [Ag(NH ]I,
4
3
)
2
which solute completely in the solution.
Comments:
● All those experiments prove the properties of silver halides, which is:
- Forming precipitate when reacts with salt.
- All kind of precipitate dissolved a little in NH OH liquid.
4
- The complex compound forming and precipitation reaction are determined by this
experiment.
● Comparing:
- The precipitation color of AgI, AgBr is more lightly, muddy than in practical
- The precipitation didn’t dissolve completely and fastly like in practical
3. Reactions of H
2
O
2
-First, we put 5 drops of 0.1 M KMnO
4
solution into a test tube.To continue, we added 5 drops of
2M H
2
SO
4
and then added 5 drops of 3 % H
2
O
2
solution into the test tube above. Finally, we
observed the change in the color and the release of gas
-With a new test tube which contain 5 drops of 0.1 M KI solution, we continued to add 5 drops of 2
M H
2
SO
4
to acidify the KI solution above. Later on, we added 5 drops of 3 % H
2
O
2
solution and had
an observation.
-In the third reaction, we added a “pinch” of solid MnO
2
into 1 mL of 3 % H
2
O
2
solution. Then, we
observed the released gas from the tube.
Reaction
Observation
Chemical Equation
Image
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8
0.1M KMnO
4
+ 2M H
2
SO
4
+ H
2
O
2
-The violet color of
KMnO
4
did not change
when H
2
SO
4
was
added.
-After adding H
2
O
2
,
the solution changed
into transparent. The
gas (as bubbles) and
heat were released.
2KMnO SO
4
+ 3H
2
4
+ 5H
2
O
2
→
8H
2
O + 2MnSO
4
+ 5O
2
+ K
2
SO
4
Explanation:
*Due to the reducing properties of H
2
O
2
; therefore, when reacting with a strong oxidizer like KMnO
4
in the acidic solution, H
2
O
2
is oxidized to Oxygen O
2
which released gas as bubbles, and KMnO
4
loses its color. So, the solution is transparent (K and MnSO is non-color).
2
SO
4
4
*In this reaction, oxygen in H
2
O
2
has oxidation number -1 is oxidized to 0 in O
2
, and Mn (in KMnO
4
)
+7 is reduced to Mn +2.
0.1M KI
+ 2M H
2
SO
4
+ H
2
O
2
-After adding H
2
O
2
,
the transparent
solution quickly
changed into
orange-yellow color,
and a dark purple
precipitate formed.
2KI + H + H
2
SO
4
2
O
2
K
2
SO
4
→
+ 2H
2
O + I
2
Explanation:
*Because H
2
O
2
has oxidizing properties; therefore, when reacting with KI
-
a reducing agent in the
acidic solution, O
-1
is reduced to O
-2
in KOH, then KOH reacts with H
2
SO SO
4
to create K
2
4
. KI is
oxidized to I
2
, which makes the orange-yellow solution and precipitate is created.
*In this reaction, oxygen in H
2
O
2
with oxidation number -1 is reduced to -2, and iodide I (in KI) -1 is
oxidized to 0 in I
2
.
H O
2 2
+ MnO
2
-After adding MnO
2
,
the gas and heat were
released. The black
solid did not dissolve.
2H
2
O
2
O + 2H O→
2
2
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Explanation:
*MnO
2
is a catalyst.
*H
2
O
2
is unstable and easy to be decomposed. When adding a “pinch” of MnO
2
to the Hydrogen
peroxide (H
2
O
2
) solution, H
2
O
2
is decomposed, with MnO
2
as a catalyst, forming H
2
O and releases
the gas – Oxygen (O
2
) (Self-redox reaction). Besides, it is also an exothermic reaction due to the heat
released.
Comments:
● From the table above, we can see that H
2
O
2
has oxidizing properties also reducing properties,
and easy to be decomposed. When reacts with different substance involved, H
2
O
2
represents
different chemical properties of it.
● The last reaction is a self-redox reaction. This reaction also let us know that MnO is a catalyst.
2
● All reactions help us understand clearly about chemical properties of H
2
O
2
, the self-redox
reaction and learn more new chemical equations.
4. Reactions of Nitrate
-At first, we added 1 mL of saturated FeSO
4
into one test tube and 1 mL of 1M NaNO
3
solution was
added into another.
-Next, we poured concentrated sulfuric acid (96%) slowly and carefully down the inside wall of the
test tube. After a few seconds, we recorded the change of color at the interface between the nitrate
solution and the concentrated sulfuric acid.
-In the second stage, we replaced FeSO by NaNO .
4
3
Reaction
Observation
Chemical Equation
Image
1M NaNO
3
+ FeSO
4
+ concentrated H
2
SO
4
-The heat and reddish
brown gas were
released.
-The transparent
solution changed to
brownish yellow.
2NaNO
3
+ 6FeSO
4
+ 4
H
2
SO
4
Na
2
SO
4
+ →
2NO + 3Fe
2
(SO
4
)
3
+
4H
2
O
2NO + O
2
2NO
2
→
Explanation: *This is a redox reaction:
*Iron (II) sulfate (FeSO
4
), Sulfuric acid (H
2
SO
4
)
and Sodium nitrate (NaNO
3
) are dissociated
completely to ions:
FeSO
4
→ Fe
2+
+ SO
4
2-
H
2
SO
4
→ 2H
+
+ SO
4
2-
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10
NaNO
3
→ Na
+
+ NO
3
-
*In the solution, with the presence of H
+
, Fe
2+
is oxidized to Fe
3+
, and N with the oxidation number +5
in ion NO
3
- +2
, is reduced to N
in NO and NO is gas, so NO is released.
*The non-color gas released is NO, NO is oxidized by O , forming NO , which is reddish brown.
2
2
*After the reaction, Fe
2
(SO ) SO
4
3
is created and its color is brownish yellow while Na
2
4
is transparent,
so the color of the solution after reacting depends on the color of Fe
2
(SO
4
)
3
.
1M NaNO
2
+ FeSO
4
+ concentrated H
2
SO
4
-The liquid was
brown-black when
FeSO
4
was added.
-After adding 96%
H
2
SO
4
, the heat and
reddish brown gas were
released, and the color
of liquid changed into
brownish yellow.
* 2NaNO
2
+ 2FeSO
4
+
3H
2
SO
4
2NaHSO
4
→
+ Fe
2
(SO
4
)
3
+ 2NO +
2H
2
O
* 2NO + O
2
2NO
2
→
Explanation: *This is also a redox reaction
*Iron (II) sulfate (FeSO
4
), Sulfuric acid (H
2
SO
4
) and Sodium nitrite (NaNO
2
) are dissociated
completely to ions:
FeSO
4
→ Fe
2+
+ SO
4
2-
H
2
SO
4
→ 2H
+
+ SO
4
2-
NaNO
2
→ Na
+
+ NO
2
-
*In the solution, Fe
2+
is oxidized to Fe
3+
, and N in NaNO
2
has the oxidation number +3, it is reduced
to +2 in NO (gas), then the gas is released.
*The non-color gas released is NO, NO is oxidized by O , forming NO , which is reddish brown.
2
2
*After the reaction, Fe
2
(SO
4
)
3
is created and its color is brownish yellow while NaHSO
4
is transparent,
so the color of the solution after reacting depends on the color of Fe .
2
(SO )
4
3
1M NaNO
3
+ FeSO
4
+concentrated
CH
3
COOH
-The solution was split
into two layers. The
upper was blood red,
and the lower was
transparent (maybe a
little yellow). In the
middle, some
precipitate was created.
* NaNO
3
+ FeSO
4
+
CH
3
COOH → NaNO
2
+ 2(CH
3
COO)
3
Fe +
2SO
3
+ 3H
2
O
* (CH
3
COO)
3
Fe +
3H
2
O → Fe(OH)
3
+
3CH
3
COOH
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11
Explanation: *This is another redox reaction.
*Iron (II) sulfate FeSO , Acetic acid COOH
4
CH
3
and Sodium nitrite NaNO
2
are dissociated completely to ions:
FeSO
4
→ Fe
2+
+ SO
4
2-
CH
3
COOH → CH
3
COO
-
+ H
+
NaNO
3
→ Na
+
+ NO
3
-
*In the solution, Fe
2+
is oxidized to Fe
3+
, and N in NaNO
3
has the oxidation number +5, it is reduced
to +3 in NaNO
2
.
*The color of the upper layer is from (CH
3
COO)
3
Fe. The lower one may be from transparent
solutions: CH
3
COOH, NaNO
2
… influenced the color of the upper layer, so it has a light color.
*The precipitate is formed because (CH
3
COO)
3
Fe is not stable, it decomposes in solution to Fe(OH)
3
(the reddish brown precipitate ).
Comments:
● From the table above, we can see when NaNO
3
/NaNO
2
reacts with FeSO
4
and concentrated
H
2
SO SO
4
, the observation is similar. The reason is that concentrated H
2
4
is a strong oxidizing
agent, so it can oxidize ions to the highest oxidation number. Furthermore, the substances in
these reactions do not change concentration and necessary volume to use, thus some products
are the same.
● The last reaction is redox. It shows us that Fe is not stable. (CH COO)
3
3
● All reactions help us understand clearly about redox reaction and learn more new chemical
equations.
5. Reactions of KMnO
4
-Prepared 3 clean test tubes and labeled # 1-3.
-In the test tube # 1: we had a combination of 10 drops of 0.5 M Na2SO3and 5 drops of 2 M
H2SO4.
-In the test tube # 2: we had a combination of 10 drops of 0.5 M Na 2SO3and 5 drops of 6N NaOH.
-In the test tube # 3: we had a combination of 10 drops of 0.5 M Na2SO3and 5 drops distilled
water.
-In the final step, we added 5 drops of 0.1M KMnO4 to each of the test tubes.
Reaction
Observation
Chemical Equation
Image
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GENERAL CHEMISTRY LABORATORY
12
0.5M Na
2
SO
3
+ 2M H
2
SO
4
+ 0.1M KMnO
4
-The color of KMnO
4
disappeared.
2KMnO SO SO 2MnSO
4
+ 3H
2
4
+ 5Na
2
3
4
+ →
K
2
SO SO
4
+ 3H O + 5Na
2
2
4
Explanation:
* In Na , S has the oxidation number +4, is oxidized by the strong oxidizer Potassium
2
SO
3
permanganate KMnO
4
in which Mn has the oxidation number +7, to form S with the oxidation number +6
and Mn with the oxidation number +2 in MnSO
4
in the acidic environment (H
2
SO
4
). So the color of turns
into non_color
0.5M Na
2
SO
3
+ 6N NaOH
+ 0.1M KMnO
4
-The solution turns
into a green-brown
solution and after a
few seconds, it turns
into a brown solution.
2KMnO SO MnO
4
+ 2NaOH + Na
2
3
→ K
2
4
+
Na
2
MnO
4
+ Na
2
SO
4
+ H
2
O
(l)
Explanation:
* In Na , S has the oxidation number +4, is oxidized by the strong oxidizer Potassium
2
SO
3
permanganate KMnO
4
in which Mn has the oxidation number +7, to form S with the oxidation number +6
and Mn with the oxidation number +6 in the basic environment. Mn +6 is in the form of Potassium
manganate K
2
MnO
4
and Sodium manganate Na
2
MnO
4
, which make the solution has the green-brown color.
After a few seconds, adding more K create MnO – a brown solution.
2
MnO
4
4
0.5M Na
2
SO
3
+ H
2
O
+ 0.1M KMnO
4
-The dark brown
precipitate appears,
and the non-color
solution occurs when
Potassium
permanganate
(KMnO
4
) loses its
color.
2KMnO SO
4
+ 3Na
2
3
+ H
2
O → 2MnO
2
+
3Na
2
SO
4
+ 2KOH
Explanation:
*In Na , S has the oxidation number +4, is oxidized by the strong oxidizer Potassium
2
SO
3
permanganate (KMnO
4
) in which Mn has the oxidation number +7, to form S with the oxidation number +6
and Mn with the oxidation number +4 in the neutral environment.
*Mn +6 is in the form of Manganese dioxide MnO which is a dark brown precipitate.
2
Comments:
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GENERAL CHEMISTRY LABORATORY
13
● All three equations are redox reaction, which changes the oxidation number of 1 or more
elements.
● In the second experiment, in fact when adding more K
2
MnO
4
that create MnO
4
– brown
precipitate, but we did not add enough K
2
MnO
4
so the solution became the brown solution
instead of the brown precipitate.
6. Reactions of Potassium Dichromate (K )
2
Cr O
2 7
-First, we added 10 drops of 0.5M K into a test tube.
2
Cr
2
O
7
-Second, we added 10 drops of 6M H and 5 drops of C OH was added at the end.
2
SO
4
2
H
5
Reaction
Observation
Image
2M K
2
Cr O
2 7
+ 6M H
2
SO
4
+ C
2
H
5
OH
-When we put H
2
SO
4
into K
2
Cr
2
O
7
, the
color of the solution
is orange.
-After we added
C
2
H
5
OH to the test
tube, the color of the
solution is dark
green.
Comments: In C
2
H
5
OH, C has the oxidation number -1, is oxidized by the strong oxidizer
Potassium dichromate K
2
Cr
2
O
7
– in which Cr has the oxidation number +6, to form C with the
oxidation number +1 and Cr with the oxidation number +3. The dark green color of the solution
comes from the color of the salt Cr Cr .
3+
2
(SO
4
)
3
7. A. Reactions of Fe
3+
-In the first step, we added 1 mL of 0.5M FeCl solution in each of the four test tubes.
3
Semester II: 2018-2019
International University, Vietnam National University - HCMC
GENERAL CHEMISTRY LABORATORY
14
-Later 5 drops of the following four reagents: 0.1M KSCN; 2M KOH;
0.5M K ]; 2M NH OH was added to four test tubes which contained FeCl solution in turn.
4
[Fe(CN)
6
4
3
-At the end of the reaction, we had observations
Reaction
Observation
Chemical Equation
Image
0.5M FeCl
3
+ 0.1M KSCN
-The solution changed
the color into blood
color (red-brown
color).
FeCl [Fe(SCN)
3
+ 3KSCN→ K
3
6
] +
3KCl
Explanation:
FeCl [Fe(SCN)
3
reacts with potassium thiocyanate (KSCN) to form the complex compound K
3
6
],
which is a red-brown solution.
0.5M FeCl
3
+ 2M KOH
-A brownish red
precipitate was
formed.
FeCl
3
+ 3KOH → Fe(OH) + 3KCl
3
Explanation:
*FeCl
3
and KOH dissociate in the solution to create ions Fe
3+
and OH
-
, then these two ions combine
together to form brownish-red precipitate Fe(OH) .
3
FeCl
3
→ Fe
3+
+ 3Cl
-
KOH → K
+
+ OH
-
Fe
3+
+ 3OH
-
→ Fe(OH)
3
↓ ⇒
0.5M FeCl
3
+ 0.5M K
4
[Fe(CN)
6
]
-Forming a dark blue
precipitate.
2FeCl [Fe(CN)
3
+ 3K
4
6
] →
Fe
4
[Fe(CN)
6
]
3
+ 12KCl
Explanation:
FeCl [Fe(CN)
3
reacts with potassium ferrocyanide (K
4
6
]) to form a complex compound Iron(III)
ferrocyanide (Fe
4
[Fe(CN)
6
]
3
), which is dark blue precipitate.
Semester II: 2018-2019
International University, Vietnam National University - HCMC
GENERAL CHEMISTRY LABORATORY
15
0.5M FeCl
3
+ 2M NH
4
OH
-The orange
precipitate appeared.
FeCl
3
+ 3NH
4
OH → Fe(OH) +
3
3NH
4
Cl
Explanation:
*FeCl
3
and NH
4
OH dissociate in the solution to create ions Fe and OH
3+
-
, then these two ions combine
together to form a light brownish red precipitate.
FeCl
3
→ Fe
3+
+ 3Cl
-
NH
4
OH → NH
4
+
+ OH
-
Fe
3+
+ 3OH
-
→ Fe(OH)
3
↓ ⇒
Comments:
● In fact, Fe(OH)
3
has a brownish red color. But experiment 5 has different color precipitate. In
experiment 5, it creates light brownish red color because of NH OH
4
is a weak base.
● Furthermore, we can learn more about the new chemical equations.
7. B. Reactions of Fe
2+
-In the second stage, we did the same as what we had done before but FeSO
4
was used instead of
FeCl
3
.
Reaction
Observation
Chemical Equation
Image
0.5M FeSO
4
+ 0.1M KSCN
-The solution changed
color into red-brown.
FeSO [Fe(SCN)
4
+ 2KSCN → K
4
6
]
+ K
2
SO
4
Explanation:
FeSO [Fe(SCN)
4
reacts with potassium thiocyanate KSCN to form the complex compound K
4
6
], which
is a red brown solution.
0.5M FeSO
4
+ 2M KOH
-A dark green
precipitate appeared.
FeSO
4
+ 2KOH
→ Fe(OH)
2
+
K
2
SO
4
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CHEMISTRY LABORATORY REPORT
EXPERIMENT 1: CHEMICAL REACTIONS Group: 4 Section: 1910 Group members: Lê Nguyễn Vân Anh Hoàng Nguyễn Minh Châu Lê Thanh Như Quỳnh Trần Băng Châu Tạ Thị Minh Thu Date: February 18th, 2019. Score:
International University, Vietnam National University - HCMC 2 GENERAL CHEMISTRY LABORATORY I.Introduction:
Chemical changes result in the formation of new substances. When a chemical reaction
occurs, substances called reactants are transformed into different substances called products that
often have different appearances and different properties. Observable signs of chemical reactions
can be a change in color, the formation o f a solid, t he re lease o f g as, a nd t he p roduction o f h eat a nd
light. We also learn how to classify chemical reactions. One classification system involves five general types of reactions: s ynthesis, d
ecomposition, single displacement, double displacement, and
combustion. Chemical reaction always happens in the living system, typically the formation and transformation of matter. II.Learning Objective:
-To perform different types of chemical reactions (acid-base, precipitate, gas forming,
complex compound forming, and oxidation-reduction reactions).
-To identify some of the products in these reactions and describe the chemical changes.
-To write and balance the chemical equations for the reactions observed. III.Materials and Equipment: Equipment Test tubes Alcohol lamp Test tube rack Looped wire Test tube holders Distilled water bottle Beakers Materials Semester II: 2018-2019
International University, Vietnam National University - HCMC 3 GENERAL CHEMISTRY LABORATORY 0.5M CuSO Concentrated CH COOH 4 3 2M NaOH 0.5M Na SO 2 3 6M NaOH 0.1M KSCN 0.1M AgNO 2M KOH 3 0.5M KBr 0.5M K [Fe(CN) ] 4 6 0.5M KI 0.5M FeSO4 0.1M KI 0.5M FeCl3 2M NH OH 0.5M Al (SO ) 4 2 4 3 3% H O 2M HCl 2 2 2M H SO 0.5M LiCl 2 4 MnO 0.5M NaCl 2 Saturated FeSO 0.5M KCl 4 0.1M NaNO 0.5M CaCl 3 2 0.1M NaNO 0.5M BaCl 2 2 0.1M KMnO4 0.5M K Cr O 2 2 7 C H OH 96% H SO 2 5 2 4 2M H SO 2 4
IV. Experimental Procedure: Data and Observations: 1. Reactions of Cu2+
-First, put 10 drops of 0.5M CuSO into each of three test tubes. 4 -Then added 10 drops of 2M N aOH i nto t he fi rst test t ube; 2 M N H OH i nto t he s econd t est t ube; 4
0.5M K [Fe(CN) ] into the third one. 4 6 - A fe w s econds l ater, added more 1 0 d rops of 2 M NaOH into t he fi
rst test t ube; 2M NH OH into 4 the second test tube. Reaction Observation Chemical Equation Image 0.5M CuSO
-The blue precipitate CuSO + 2 NaOH → Cu(OH) 4 4  2 + 2M NaOH and colloidal + Na SO 2 4 phenomena were formed Semester II: 2018-2019
International University, Vietnam National University - HCMC 4 GENERAL CHEMISTRY LABORATORY Explanation:
* CuSO is dissociated completely: 4
CuSO → Cu2++ SO 2- 4  4
* NaOH is also dissociated completely to form Hydroxide ion OH-:
NaOH →  Na+ + OH-
*Ion Cu2+and OH-in the solution combine together to form Cu(OH) , a blue precipitate. 2  0.5M CuSO -First, the blue
CuSO + 2NH OH → Cu(OH) + 4 4  4 2 + 2M NH OH precipitate dissolved (NH ) SO 4 4 2 4 after being formed. Then, the solution Cu(OH) + 4NH 2 →  3  turned into dark blue. [Cu(NH ) ](OH) 3 4 2 Explanation:
* CuSO is dissociated completely: 4 CuSO 2+  2-. 4 →  Cu + SO  4
* NH OH is also dissociated completely: 4
NH OH → NH4++ OH- 4
*Ion Cu2+and OH-in the solution combine together to form blue precipitate Cu(OH) because it has 2 very low solubility.
*Then, the remaining NH OH reacts with the precipitate Cu(OH) to form a 4 2
complex-compound [Cu(NH ) ]SO4 to form the dark blue of the solution. 3  4 0.5M CuSO -The
brown 2CuSO +K Fe(CN) → 4 4 4 6 0.5M K [Fe(CN) ] precipitate
and Cu [Fe(CN) ] + 2K SO 4 6 2 6 2 4 colloidal phenomena were formed. Explanation:
* Copper (II) sulfate CuSO is dissociated completely: 4 
CuSO → Cu2+ + SO42- 4  Comments:
● The feature of Cu2+ is it can react with OH- form blue precipitation.
● The solution Cu(OH) also react with NH OH to form a c omplex c ompound s o reaction 2 has 2 4 the dark blue at the end.
● The last precipitate solution has brown color because the solution is compound consist of Fe2+
● Double displacement reaction occurs when part of one reactant is replaced by part of another reactant. Semester II: 2018-2019
International University, Vietnam National University - HCMC 5 GENERAL CHEMISTRY LABORATORY ● Comparing: -
The color of second precipitate is less dark than in practice. -
The color of third precipitate is just only brown unlike red-brown in practical.
2. Reactions of silver halides
-Prepared 0.5 M solutions of KCl, KBr, and KI.
-Then added 10 drops of 0.1M silver nitrate to 10 drops of each salt solution: KCl, KBr, and KI. Observing the result
-Next, we divided each of the solutions equally into two test tubes labeled # 1-3 A and # 1-3 B
*With test tube #1-3 A: did not add anything, just observe.
*With test tube # 1-3 B: added 5 drops of 2M NH OH to each of test tubes. 4
-Waited till the end of the solution and recorded the observation. Reaction Observation Chemical Equation Image 0.5M KCl
-The white precipitate AgNO + KCl → AgCl + KNO 3 3 + 0.1M AgNO was formed. 3 Explanation:
*KCl dissociate completely to form Cl-
*In the solution, ion Cl- combine with ion Ag+ to fo rm A
gCl which i s a white precipitate because they have low solubility. 0.5M KCl
-The white precipitate AgNO + KCl → AgCl + KNO 3 3 + 0.1M AgNO was formed. 3 + 2M NH OH -After adding NH OH, AgCl + 2NH OH → 4 4 4
the precipitate was [Ag(NH ) ]Cl + H O 3 2  2 dissolved slowly. Explanation:
*The precipitate AgCl reacts with Ammonium hydroxide NH OH to create the complex – compound 4
[Ag(NH ) ]Cl, which solute completely in the solution. 3  2 Semester II: 2018-2019
International University, Vietnam National University - HCMC 6 GENERAL CHEMISTRY LABORATORY 0.5M KBr -The light yellow AgNO + KBr →AgBr + KNO 3 3 + 0.1M AgNO precipitate was 3 formed. Explanation:
*KBr dissociate completely to form BRr-
*In the solution, ion Br- combine with ion Ag+
to form AgBr which is a white-yellow precipitate because they have low solubility. 0.5M KBr -The light yellow AgNO +KBr → AgBr + KNO 3 3 + 0.1M AgNO precipitate was 3 + 2M NH OH formed. AgBr + 2NH OH → 4 4
-After adding NH OH, [Ag(NH ) ]Br + H O 4 3 2 2 the precipitate was dissolved a little. Explanation:
*The precipitate AgBr reacts with NH OH to create the complex-compound [Ag(NH ) ]Br, which 4 3 2
solute completely in the solution. 0.5M KI
-The milky- white AgNO + KI → AgI + KNO 3 3 + 0.1M AgNO yellow. 3 Explanation:
*KI dissociate completely to form I-
*In the solution, ion I- combine with ion Ag+
to form AgI which is a light yellow precipitate because they have low solubility. Semester II: 2018-2019
International University, Vietnam National University - HCMC 7 GENERAL CHEMISTRY LABORATORY 0.5M KI -The milky- white
AgI+2NH OH → [Ag(NH ) ]I 4 3 2 + 0.1M AgNO3 yellow precipitate + H O 2 + 2M NH OH appeared. 4 -After adding NH OH, 4 the precipitate dissolved hardly. Explanation:
*The precipitate AgI reacts with NH OH to create a milky yellow complex – compound [Ag(NH ) ]I, 4 3 2
which solute completely in the solution. Comments:
● All those experiments prove the properties of silver halides, which is: -
Forming precipitate when reacts with salt. -
All kind of precipitate dissolved a little in NH OH liquid. 4 -
The complex compound forming and precipitation reaction are determined by this experiment. ● Comparing: -
The precipitation color of AgI, AgBr is more lightly, muddy than in practical -
The precipitation didn’t dissolve completely and fastly like in practical 3. Reactions of H O 2 2
-First, we put 5 drops of 0.1 M KMnO solution into a test tube.To continue, we added 5 drops of 4
2M H SO and then added 5 drops of 3 % H O solution into the test tube above. Finally, we 2 4 2 2
observed the change in the color and the release of gas -With a new test tube which c ontain 5 d rops of 0 .1 M K I s olution, w e c ontinued t o add 5 drops o f 2
M H SO t o acidify the KI solution a bove. Later o n, w e a dded 5 drops o f 3 % H O solution a nd had 2 4 2 2  an observation.
-In the third reaction, we added a “pinch” of solid Mn O i nto 1 mL o f 3 % H O s olution. T hen, w e 2 2 2
observed the released gas from the tube. Reaction Observation Chemical Equation Image Semester II: 2018-2019
International University, Vietnam National University - HCMC 8 GENERAL CHEMISTRY LABORATORY 0.1M KMnO
-The violet color of 2KMnO + 3H SO + 5H O → 4 4 2 4 2 2 + 2M H SO KMnO did n
ot change 8H O + 2MnSO + 5O + K SO 2 4 4 2 4 2 2 4 + H O when H SO was 2 2 2 4 added. -After adding H O , 2 2 the solution changed into transparent. The gas (as bubbles) and heat were released. Explanation:
*Due to the reducing properties of H O ; therefore, when reacting with a strong oxidizer like KMnO 2 2 4
in the acidic solution, H O is oxidized to Oxygen O which released gas as bubbles, and KMnO 2 2 2 4
loses its color. So, the solution is transparent (K SO and MnSO is non-color). 2 4 4 *In this reaction, oxygen in H  O h as o xidation number -1 i s o xidized t o 0 i n O  , and Mn (in K MnO ) 2 2 2 4 +7 is reduced to Mn +2. 0.1M KI
-After adding H O , 2KI + H  SO + H  O → K SO 2 2 2 4 2 2 2 4 + 2M H SO the transparent + 2H O + I 2 4 2 2 + H O solution quickly 2 2 changed into orange-yellow color, and a dark purple precipitate formed. Explanation:
*Because H O has oxidizing properties; therefore, when reacting with KI- a reducing agent in the 2 2
acidic solution, O-1 is reduced to O-2 in KOH, then KOH reacts with H SO to create K SO . KI is 2 4 2 4
oxidized to I , which makes the orange-yellow solution and precipitate is created. 2
*In this reaction, oxygen in H O with oxidation number -1 i s reduced t o -2 , and iodide I (in KI) -1 is 2 2 oxidized to 0 in I . 2 H O
-After adding MnO , 2H O → O + 2H O 2 2 2 2 2 2 2 + MnO the gas and heat were 2 released. The black solid did not dissolve. Semester II: 2018-2019
International University, Vietnam National University - HCMC 9 GENERAL CHEMISTRY LABORATORY Explanation: *MnO is a catalyst. 2
*H O is unstable and easy to be decomposed. When adding a “pinch” of MnO to the Hydrogen 2 2 2
peroxide (H O ) solution, H O is decomposed, with MnO as a catalyst, forming H O and releases 2 2 2 2 2 2
the gas – Oxygen (O ) (Self-redox reaction). Besides, it is also an e xothermic reaction d ue t o the heat 2 released. Comments:
● From the table above, we can see that H O has oxidizing properties also reducing properties, 2 2
and easy to be decomposed. When reacts with different substance involved, H O represents 2 2
different chemical properties of it.
● The last reaction is a self-redox reaction. This reaction also let us know that MnO is a catalyst. 2
● All reactions help us understand clearly about chemical properties of H O , the self-redox 2 2
reaction and learn more new chemical equations. 4. Reactions of Nitrate -At first, we added 1 mL o f saturated Fe SO into o ne t est t ube a nd 1 mL of 1 M NaNO s olution w as 4  3 added into another.
-Next, we poured concentrated sulfuric acid (96%) s lowly a nd c arefully d own the inside wall of t he
test tube. After a few seconds, we recorded the change of color at the interface b etween the nitrate
solution and the concentrated sulfuric acid.
-In the second stage, we replaced FeSO by NaNO . 4 3 Reaction Observation Chemical Equation Image 1M NaNO -The heat and reddish 3 2NaNO + 6FeSO + 4 + FeSO brown gas were 3  4 4 H SO → Na SO + + concentrated H SO released. 2 4 2 4  2 4 -The
transparent 2NO + 3Fe (SO ) + 2 4 3 solution changed to 4H O 2 brownish yellow. 2NO + O → 2NO 2 2
Explanation: *This is a redox reaction:
*Iron (II) sulfate (FeSO ), Sulfuric acid (H SO ) and Sodium nitrate (NaNO ) are dissociated 4 2 4  3 completely to ions:
FeSO → Fe2+ + SO 2- 4 4
H SO → 2H+ + SO 2- 2 4 4 Semester II: 2018-2019
International University, Vietnam National University - HCMC 10 GENERAL CHEMISTRY LABORATORY
NaNO → Na+ + NO  - 3 3
*In the solution, with the presence o f H+, Fe 2+ is o
xidized t o Fe3+, and N w ith t he o xidation n umber + 5 in ion NO - +2
, is reduced to N  in NO and NO is gas, so NO is released. 3
*The non-color gas released is NO, NO is oxidized by O , forming NO , which is reddish brown. 2 2
*After the reaction, Fe (SO ) is created and its color is b rownish y ellow w hile N a SO is t ransparent, 2 4 3 2 4 
so the color of the solution after reacting depends on the color of Fe (SO ) . 2 4 3 1M NaNO -The liquid was * 2NaNO + 2FeSO + 2 2  4 + FeSO brown-black
when 3H SO → 2NaHSO 4 2 4 4 + concentrated H SO FeSO was added. + Fe (SO ) + 2NO + 2 4 4 2 4 3 -After adding 96% 2H O 2 H SO , the heat and 2 * 2NO + O → 2NO  4 2 2 reddish brown gas were released, and the color of liquid changed into brownish yellow.
Explanation: *This is also a redox reaction
*Iron (II) sulfate (FeSO ), Sulfuric acid (H SO ) and Sodium nitrite (NaNO ) are dissociated 4 2 4 2 completely to ions:
FeSO → Fe2+ + SO 2- 4 4
H SO → 2H+ + SO 2- 2 4 4
NaNO → Na+ + NO  - 2 2
*In the solution, Fe2+ is oxidized to Fe3+, and N in NaNO has the oxidation number + 3, it is reduced 2
to +2 in NO (gas), then the gas is released.
*The non-color gas released is NO, NO is oxidized by O , forming NO , which is reddish brown. 2 2
*After the reaction, Fe (SO ) i s c reated a nd its color i s b rownish y ellow w hile N aHSO is t ransparent, 2 4 3 4 
so the color of the solution after reacting depends on the color of Fe (SO ) . 2 4 3 1M NaNO
-The solution was split * NaNO + FeSO + 3 3 4 + FeSO
into two layers. The CH COOH → NaNO 4 3 2 +concentrated
upper was blood red, + 2(CH COO) Fe + 3 3 CH COOH
and the lower was 2SO + 3H O 3 3  2 transparent (maybe a * (CH COO) Fe + 3 3
little yellow). In the 3H O → Fe(OH) + 2 3 middle, some 3CH COOH 3 precipitate was created. Semester II: 2018-2019
International University, Vietnam National University - HCMC 11 GENERAL CHEMISTRY LABORATORY
Explanation: *This is another redox reaction.
*Iron (II) sulfate FeSO , Acetic acid CH COOH 4 3
and Sodium nitrite NaNO are dissociated completely to ions: 2
FeSO → Fe2+ + SO 2- 4 4
CH COOH → CH COO- + H+ 3 3
NaNO → Na+ + NO  - 3 3
*In the solution, Fe2+ is oxidized to Fe3+, and N in NaNO has the oxidation number + 5, it is reduced 3 to +3 in NaNO . 2
*The color of the upper layer is from (CH COO) Fe. The lower one may be from transparent 3 3
solutions: CH COOH, NaNO … influenced the color of the upper layer, so it has a light color. 3 2
*The precipitate is formed because (CH COO) Fe is not stable, it decomposes in solution to Fe(OH) 3 3 3
(the reddish brown precipitate ). Comments:
● From the table above, we can see when NaNO /NaNO reacts with FeSO and concentrated 3  2 4
H SO , the observation is similar. The reason is that concentrated H SO is a strong oxidizing 2 4 2 4
agent, so it can oxidize ions to the highest oxidation number. Furthermore, the substances in
these reactions do not change concentration and necessary volume to use, thus some products are the same.
● The last reaction is redox. It shows us that (CH COO) Fe is not stable. 3 3
● All reactions help us understand clearly about redox reaction and learn more new chemical equations. 5. Reactions of KMnO4
-Prepared 3 clean test tubes and labeled # 1-3.
-In the test tube # 1: we had a combination of 10 drops of 0.5 M Na2SO3and 5 drops of 2 M H2SO4.
-In the test tube # 2: we had a combination of 10 drops of 0.5 M Na 2SO3and 5 drops of 6N NaOH.
-In the test tube # 3: we had a combination of 10 drops of 0.5 M Na2SO3and 5 drops distilled water.
-In the final step, we added 5 drops of 0.1M KMnO4 to each of the test tubes. Reaction Observation Chemical Equation Image Semester II: 2018-2019
International University, Vietnam National University - HCMC 12 GENERAL CHEMISTRY LABORATORY 0.5M Na SO -The color of KMnO
2KMnO + 3H SO + 5Na SO → 2MnSO + 2 3 4 4 2 4 2 3 4 + 2M H SO disappeared.
K SO + 3H O + 5Na SO 2 4 2 4 2 2 4 + 0.1M KMnO 4 Explanation:
* In Na SO , S has the oxidation number +4, is oxidized by the strong oxidizer Potassium 2 3
permanganate KMnO in which Mn has the oxidation number +7, to form S with the oxidation number +6 4
and Mn with the oxidation number +2 in MnSO in the acidic environment (H SO ). So the color of turns 4 2 4 into non_color 0.5M Na SO
-The solution turns 2KMnO + 2NaOH + Na SO → K MnO + 2 3 4 2 3 2 4 + 6N NaOH
into a green-brown Na MnO + Na SO + H O 2 4 2 4  2 (l) + 0.1M KMnO solution and after a 4 few seconds, it turns into a brown solution. Explanation:
* In Na SO , S has the oxidation number +4, is oxidized by the strong oxidizer Potassium 2 3
permanganate KMnO in which Mn has the oxidation number +7, to form S with the oxidation number + 6 4
and Mn with the oxidation number +6 in the basic environment. Mn +6 is in the form of Potassium
manganate K MnO and Sodium manganate Na MnO , which make the solution has the green-brown color. 2 4  2 4
After a few seconds, adding more K MnO create MnO – a brown solution. 2 4 4 0.5M Na SO -The dark
brown 2KMnO + 3Na SO + H O → 2MnO + 2 3 4 2 3 2 2  + H O precipitate appears, 3Na SO + 2KOH 2 2 4 + 0.1M KMnO and the non-color 4 solution occurs when Potassium permanganate (KMnO ) loses its 4 color. Explanation:
*In Na SO , S has the oxidation number +4, is oxidized by the strong oxidizer Potassium 2 3
permanganate (KMnO ) in which Mn has the oxidation number +7, t o fo rm S with t he o xidation number + 6 4
and Mn with the oxidation number +4 in the neutral environment.
*Mn +6 is in the form of Manganese dioxide MnO which is a dark brown precipitate. 2 Comments: Semester II: 2018-2019
International University, Vietnam National University - HCMC 13 GENERAL CHEMISTRY LABORATORY
● All three equations are redox reaction, which changes the oxidation number of 1 or more elements.
● In the second experiment, in fact when adding more K MnO that create MnO – brown 2 4 4
precipitate, but we did not add enough K MnO so the solution became the brown solution 2 4
instead of the brown precipitate.
6. Reactions of Potassium Dichromate (K Cr O ) 2 2 7
-First, we added 10 drops of 0.5M K Cr O into a test tube. 2 2 7
-Second, we added 10 drops of 6M H SO and 5 drops of C H OH was added at the end. 2 4 2 5 Reaction Observation Chemical Equation Image 2M K Cr O
-When we put H SO 2K Cr O + 3C H OH + 8H SO 2 2 7 2 4 2 2 7 2 5 2 4 + 6M H SO
into K Cr O , the → 3CH COOH + 2Cr (SO ) + 2 4 2 2 7 3 2 4 3 + C H OH
color of the solution 2K SO + 11H O 2 5 2 4 2 is orange. -After we added C H OH to the test 2 5 tube, the color of the solution is dark green.
Comments: In C H OH, C has the oxidation number -1, is oxidized by the strong oxidizer 2 5
Potassium dichromate K Cr O – in which Cr has the oxidation number +6, to form C with the 2 2 7 
oxidation number +1 and Cr with the oxidation number +3. The dark green color of the solution
comes from the color of the salt Cr3+ Cr (SO ) . 2 4 3 7. A. Reactions of Fe3+
-In the first step, we added 1 mL of 0.5M FeCl solution in each of the four test tubes. 3 Semester II: 2018-2019
International University, Vietnam National University - HCMC 14 GENERAL CHEMISTRY LABORATORY
-Later 5 drops of the following four reagents: 0.1M KSCN; 2M KOH;
0.5M K [Fe(CN) ]; 2M NH OH was added to four test tubes which contained FeCl solution in turn. 4 6 4 3
-At the end of the reaction, we had observations Reaction Observation Chemical Equation Image 0.5M FeCl -The solution c
hanged FeCl + 3KSCN→ K [Fe(SCN) ] + 3 3 3 6 + 0.1M KSCN the color into blood 3KCl color (red-brown color). Explanation:
FeCl reacts with potassium thiocyanate (KSCN) to form the complex compound K [Fe(SCN) ], 3 3 6
which is a red-brown solution. 0.5M FeCl -A brownish
red FeCl + 3KOH → Fe(OH) + 3KCl 3 3 3 + 2M KOH precipitate was formed. Explanation:
*FeCl and KOH dissociate in the solution to create ions Fe3+and OH-, then these two ions combine 3
together to form brownish-red precipitate Fe(OH) . 3
FeCl → Fe3+ + 3Cl- 3  KOH → K+ + OH- ⇒ F
e3+ + 3OH-→ Fe(OH) ↓ 3 0.5M FeCl -Forming a dark blue 2FeCl + 3K [Fe(CN) ] → 3 3 4 6
+ 0.5M K [Fe(CN) ] precipitate. Fe [Fe(CN) ] + 12KCl 4 6 4 6 3 Explanation:
FeCl reacts with potassium ferrocyanide (K [Fe(CN) ]) to form a complex compound Iron(III) 3 4 6
ferrocyanide (Fe [Fe(CN) ] ), which is dark blue precipitate. 4 6 3 Semester II: 2018-2019
International University, Vietnam National University - HCMC 15 GENERAL CHEMISTRY LABORATORY 0.5M FeCl -The orange
FeCl + 3NH OH → Fe(OH) + 3 3  4 3 + 2M NH OH precipitate appeared. 3NH Cl 4 4 Explanation:
*FeCl and NH OH dissociate i n the solution t o create ions Fe3+ and OH-, then these t wo i ons combine 3 4
together to form a light brownish red precipitate.
FeCl → Fe3+ + 3Cl- 3 
NH OH → NH + + OH - 4 4 ⇒ F
e3+ + 3OH- → Fe(OH) ↓ 3 Comments:
● In fact, Fe(OH) has a brownish red color. But experiment 5 has different color precipitate. In 3 
experiment 5, it creates light brownish red color because of NH OH 4 is a weak base. 
● Furthermore, we can learn mo 
re about the new chemical equations. 7. B. Reactions of Fe2+
-In the second stage, we did the same as what we had done before but FeSO was used instead of 4 FeCl . 3 Reaction Observation Chemical Equation Image 0.5M FeSO
-The solution changed FeSO + 2KSCN → K [Fe(SCN) ] 4 4 4 6 + 0.1M KSCN color into red-brown. + K SO 2 4 Explanation:
FeSO reacts with potassium thiocyanate K SCN t o fo
rm the complex compound K [Fe(SCN) ], w hich 4 4 6 is a red brown solution. 0.5M FeSO -A dark green
FeSO + 2KOH → Fe(OH) + 4 4  2 + 2M KOH precipitate appeared. K SO 2 4 Semester II: 2018-2019