Stoichiometry of Reaction Report - IPSE FPMIPA Universitas Pendidikan Indonesia Assignment - Fundamental Chemistry
STOICHIOMETRY OF REACTION
A.
Place
and Date
Wet Lab IPSE / 22 November 2016
B.
Objectives
1.
Determine coefficient reaction based on precipitation formed.
2. Determine stoichiometry of acid-base
reaction.
C.
Basic
Theory
A chemical reaction is a process in which one set of
substances called reactants is converted to new set of substances called
products. In many cases, though, nothing happens when or substances are mixed
together, they retain their original properties and composition. Just as
symbols are used for elements and formulas for compounds, there is a symbolic
or shorthand way of representing a chemical reaction called chemical equation.
All
atoms present in the reactants must be accounted for among the products. In
other words, there must be the same number of each type of atom on the product
side and on the reactant side of the narrow. Making sure that this rule is
followed is called balancing chemical equation. The coefficients in balanced
equation tell the number of substances involved in a reaction. These
coefficients also tell the number of moles of substances in a reactions. For
example :
CH4(g) + 2O2(g) Ã
CO2(g) + 2 H2O(g)
The
equation tell us that 1 molecule of CH4 reacts with 2 molecules O2
to produce 1 molecule of CO2 and 2 molecules of H2O.
The equation also tells us that 1n
mole of CH4 reacts with 2n
moles of O2 to give 1n
mole of CO2 and 2n moles
of H2O (n = any value).
In
the laboratory, coefficient in chemical reaction can be determined through
experimental data series results. One of simple way to determine coefficient of
reaction is continuous variation method. The basic principle of this method is
in a series experiments, the total molar amount of reactants is made permanent,
while the molar amount of changed regularly. Changes that occur as a result of
the reaction between the mixture of reactants (mass, volume, and temperature)
is plotted to the molar of each reactant in a graph, in order to obtain an
optimum point. Optimum point shows coefficient ration for each reactant.
D.
Equipment
and Materials
Equipment
-
Beakers 100ml (6)
-
Graduated cylinder 50ml
(1)
-
Graduated cylinder 100ml
(1)
-
Thermometer (2)
-
Tray (1)
-
Ruler (1)
Materials
- NaOH 1 M
- NaOH 0,1 M
- CuSO4 0,1 M
- HCl 1 M
E.
Procedures
1.
Stoichiometry
of Precipitation Reaction
a. Two
beakers are provided. 5 ml NaOH 0,1 M is taken and filled in beaker 1, 25 ml
CuSO4 0,1 M is filled in beaker 2.
b. Two
kinds of solution are mixed and shaked.
c. The
mixture is allowed to form a precipitate.
d. The
high of precipitation is measured using ruler (mm).
e. The
same way is done with different composition of reactants (total volume still
30ml)
·
10 ml NaOH 0,1 M and 20
ml CuSO4 0,1 M
·
15 ml NaOH 0,1 M and 15
ml CuSO4 0,1 M
·
20 ml NaOH 0,1 M and 10
ml CuSO4 0,1 M
·
25 ml NaOH 0,1 M and 5 ml
CuSO4 0,1 M
f.
Graph is made and high of
precipitation [Y] VS volume NaOH/ CuSO4 [X] is plotted.
g. Optimum
point and coefficient of reaction are determined.
2.
Stoichiometry
of Acid-Base Reaction
a. 5
ml NaOH 1 M is taken and filled in beaker 1, 25 ml HCl 1 M is filled in beaker
2. The temperature of both solutions (TM) are measured until both of solution
has the same temperature (using water in tray).
b. Two
kinds of solution are mixed and the constant temperature (TA) is measured.
c. The
same way is done with different composition of reactants (total volume still 30
ml)
d. Graph
is made and the change of temperature [Y] VS volume NaOH/ CuSO4 [X]
is plotted.
e. Optimum
point and coefficient of reaction are determined.
F.
Observation
Table
1.
Stoichiometry
of Precipitation Reaction
|
Experiment |
Volume
of NaOH 0,1 M |
Volume
of CuSO4 0,1 M |
Precipitation
(mm) |
|
1. |
5 mL |
25 mL |
1 mm |
|
2. |
10 mL |
20 mL |
2 mm |
|
3. |
15 mL |
15 mL |
3 mm |
|
4. |
20 mL |
10 mL |
13 mm |
|
5. |
25 mL |
5 mL |
11 mm |
2.
Stoichiometry
of Acid-Base Reaction
|
Experiment |
Volume
of NaOH 1 M |
Volume
of CuSO4 1 M |
TM |
TA |
∆T |
|
1. |
5 mL |
25 mL |
24 |
27 |
3 |
|
2. |
10 mL |
20 mL |
25 |
27 |
2 |
|
3. |
15 mL |
15 mL |
24 |
29 |
5 |
|
4. |
20 mL |
10 mL |
24 |
28 |
4 |
|
5. |
25 mL |
5 mL |
24 |
25 |
1 |
G.
Chemical
Reaction and Calculation
1.
Stoichiometry
of Precipitation Reaction
|
Volume
of NaOH 0,1 M |
Volume
of CuSO4 0,1 M |
Mol of NaOH |
Mol
of CuSO4 |
|
5 mL = 0,005 L |
25 mL =0,025 L |
n = M/v |
n = M/v |
|
10 mL = 0,01 L |
20 mL = 0,02 L |
n = M/v |
n = M/v |
|
15 mL = 0,015 L |
15 mL = 0,015 L |
n = M/v |
n = M/v |
|
20 mL = 0,02 L |
10 mL = 0,01 L |
n = M/v |
n = M/v |
|
25 mL = 0,025 L |
5 mL = 0,005 L |
n = M/v |
n = M/v |
·
Reaction between 5 mL
NaOH 0,1 M with 25 ml CuSO4 0,1 M
2NaOH (aq) + CuSO4 (aq) Ã
Na2SO4 (aq) + Cu(OH)2 (s)
Initial :
20 mol 4 mol
Reactant : 8 mol 4 mol 4 mol 4 mol
Rest : 12 mol -
4 mol 4 mol
·
Reaction between 10 mL
NaOH 0,1 M with 20 mL CuSO4 0,1 M
2NaOH (aq) + CuSO4 (aq) Ã
Na2SO4 (aq) + Cu(OH)2 (s)
Initial :
10 mol 5 mol
Reactant : 10 mol 5 mol 5
mol 5 mol
Rest : -
- 5 mol 5 mol
·
Reaction between 15 mL
NaOH 0,1 M with 15 mL CuSO4 0,1 M
2NaOH (aq) + CuSO4 (aq) Ã
Na2SO4 (aq) + Cu(OH)2 (s)
Initial :
6,67 mol 6,67 mol
Reactant : 6,67 mol 3,335
mol 3,335 mol 3,335
mol
Rest : -
3,335 mol 3,335 mol 3,335
mol
·
Reaction between 20 mL
NaOH 0,1 M with 10 mL CuSO4 0,1 M
2NaOH (aq) + CuSO4 (aq) Ã
Na2SO4 (aq) + Cu(OH)2 (s)
Initial :
5 mol 10 mol
Reactant : 5 mol 2,5 mol 2,5 mol 2,5 mol
Rest : -
7,5 mol 2,5 mol 2,5 mol
·
Reaction between 25 mL
NaOH 0,1 M with 5 mL CuSO4 0,1 M
2NaOH (aq) + CuSO4 (aq) Ã
Na2SO4 (aq) + Cu(OH)2 (s)
Initial :
4 mol 20 mol
Reactant : 4 mol 2 mol
2 mol 2 mol
Rest : -
18 mol 2 mol 2
mol
2.
Stoichiometry
of Acid-Base Reaction
|
Volume
of NaOH 1 M |
Volume
of HCl 1 M |
Mol of NaOH |
Mol of HCl |
|
5 mL = 0,005 L |
25 mL =0,025 L |
n = M/v |
n = M/v |
|
10 mL = 0,01 L |
20 mL = 0,02 L |
n = M/v |
n = M/v |
|
15 mL = 0,015 L |
15 mL = 0,015 L |
n = M/v |
n = M/v |
|
20 mL = 0,02 L |
10 mL = 0,01 L |
n = M/v |
n = M/v |
|
25 mL = 0,025 L |
5 mL = 0,005 L |
n = M/v |
n = M/v |
·
Reaction between 5 mL
NaOH 1 M and 25 mL HCL 1 M
NaOH (aq) + HCl (aq) Ã
NaCl (aq) + H2O (l)
Initial : 200 mol
40 mol
Reactant : 40 mol 40
mol 40
mol 40 mol
Rest : 160
mol - 40 mol 40
mol
·
Reaction between 10 mL
NaOH 1 M and 20 mL HCL 1 M
NaOH
(aq) + HCl (aq) Ã NaCl (aq) + H2O (l)
Initial : 100 mol
50 mol
Reactant : 50 mol 50 mol
50 mol 50
mol
Rest : 50
mol - 50 mol
50 mol
·
Reaction between 15 mL
NaOH 1 M and 15 mL HCL 1 M
NaOH (aq) +
HCl (aq) Ã NaCl (aq) + H2O (l)
Initial : 66,67 mol
66,67 mol
Reactant : 66,67 mol
66,67 mol 66,67 mol
66,67 mol
Rest : - - 66,67 mol
66,67 mol
·
Reaction between 20 mL
NaOH 1 M and 10 mL HCL 1 M
NaOH
(aq) + HCl (aq) Ã NaCl (aq) + H2O (l)
Initial : 50 mol 100 mol
Reactant : 50 mol 50 mol
50 mol 50
mol
Rest : -
50 mol 50
mol 50 mol
·
Reaction between 25 mL
NaOH 1 M and 5 mL HCL 1 M
NaOH
(aq) + HCl (aq) Ã NaCl (aq) + H2O (l)
Initial : 40 mol 200 mol
Reactant : 40 mol 40 mol
40 mol 40
mol
Rest : -
160 mol 40
mol 40 mol
H.
Data
Analysis (complete with graph)
1. Stoichiometry of
Precipitation Reaction
In first reaction, we
use NaOH and CuSO4. The reaction of 2NaOH
+ CuSO4 Ã
Na2SO4 + Cu(OH)2, the SO42-
anion is joining with an element form salts, then most of the salt formed is a
soluble salt. Including when the sulfate anion reacts with Na. While hydroxide
(OH-) only dissolves in water. So that, when OH- reacts
with Cu forming Cu(OH)2 will form in soluble compounds. So, the
precipitation formed from the reaction of NaOH and CuSO4 is Cu(OH)2.
The total volume between two solutions
should be 30 mL.
Reaction between 20 mL
NaOH + 10 mL CuSO4 actually produce
the highest precipitation appropriate literature. Reaction of 5 mL solution of
NaOH 0.1 M and 25 mL of 0.1 M CuSO4 produced high of precipitation
about 1mm and the color of precipitate is light blue color. Reaction of 10 mL NaOH
0,1 M and 20 mL CuSO4 0,1 M, the high of precipitation is 2 mm, with
a light blue color. Then, reaction of 15 mL NaOH 0.1 M with 15 mL CuSO4 0,1
M, the high of precipitation is 3mm, with a light blue color. Reaction of 20 mL
NaOH 0,1 M and 10 mL CuSO4 0,1
M produces the high of precipitation is about 13 mm and it has dark
blue color in precipitate, while 25 mL NaOH 0.1 M mixed 5 mL CuSO4,
the high of precipitate is 11 mm, with black color.
From the results that we
had, the more volume of NaOH that we mixed, it will produce progressively
darker colors. If the volume of NaOH that we mixed is less, the color of
solution increasingly clear or bright. Then, the more volume of NaOH that we
mixed, the higher of sediment that we get. However, that level continues to
rise until it reaches the optimum point. And the optimum point is in reaction
of 20 mL NaOH and 10 mL CuSO4 produces 13 mm of high precipitation.
From that reaction, we got balance coefficient of NaOH and CuSO4
is 2 : 1.
2.
Stoichiometry
of Acid-Base Reaction
The second reaction is between NaOH and HCl. This reaction called neutralization.
The relation between temperature and the stoichiometry reaction is the temperature will reach a optimum point or the maximum value. The lowest temperature is called minimum point and the highest temperature is called optimum point. Usually the optimum point obtained if the reaction based on stoichiometry. In reaction between 5 mL NaOH 1 M and 25 mL HCL 1 M is non stoichiometry
reaction because HCl completely reacted first and NaOH still remain 160 mol. In
reaction between 10 mL NaOH 1 M and 20 mL HCL 1 M is non stoichiometry reaction
because HCl completely reacted first and NaOH still remain 50 mol. The reaction
between 20 mL NaOH 1 M and 10 mL HCL 1 M is non stoichiometry reaction because
NaOH completely reacted first and HCl still remain 50 mol. The reaction between 25 mL
NaOH 1 M and 5 mL HCL 1 M is non stoichiometry reaction because NaOH completelyreacted first and HCl still remain 160 mol. In reaction between 15 mL NaOH 1 M and 15
mL HCL 1 M is stoichiometry reaction because all reagents completely reacted and there was nothing left. So, we obtain the optimum point from this reaction. Not only based on stoichiometry, but also we find the optimum point from a graph of volume and temperature changes, the optimum point obtained at 15 mL NaOH 1 M and 15 mL HCL 1 M with the change in temperature of 50.
Reaction coefficients based on the optimum point is 1: 1, means one mole of NaOH reacts with one mole of HCl. The equation is based on the optimum point by equalizing equation isthe same that produced one mole of NaCl and one mole of water.
I. Conclusion
If there are two substances that are mixed, it will cause a change in temperature, color and precipitation.
Coefficient of reaction can be determined from the maximum point of a reaction. The optimum point in
the precipitation reaction occurs in a volume of 20 mL NaOH and 10 mL CuSO4 which has high precipitation is about 13 mm high. NaOH and CuSO4 reaction coefficient ratio is 2: 1.
In this case, changes in temperature did not based on volume. The optimum point in acid-base reaction system obtained in 15 mL HCl and 15 mL NaOH with changes in temperature of 5° C. Comparison of HCl and NaOH reaction coefficient is 1: 1.
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