CHAPTER FIVE
CONCLUSION AND RECOMMENDATION
5.1 Conclusion
The synthesis of zeolite from kaolin was successful as we have good crystals of supposed zeolite Y. The crucial stages of the preparation of zeolites are dehydroxylation and zeolitization. The optimum conditions for preparation of metakaolin were temperature and time of 650˚C and 60minutes respectively. Various conditions, different ageing times and crystallization temperatures, for ageing and crystallization were evaluated. The conditions of 3hours ageing time, 80˚C and 9hours crystallization temperature and time respectively appear to be the best as they produced the best zeolite crystals.
Okpella kaolin clay is a good material for development of zeolites in Nigeria.
5.2 Recommendations
Based on the issues encountered during experimentation, the following recommendations are suggested
1) Characterization of samples should be done after each stage to enhance better results.
2) Steady power supply will enhance performance and speed of reaction during operation of the furnace, ovens (autoclave reactor) and other equipment.
3) Smaller particle size during refining should be used as it provides for larger catalyst surface area due to the particle-size and surface area relationship.
4) Metakaolin could be used as a single starter material by dealuminating with acids.
5) Factorial designs should be implemented to assist with modeling of the zeolitization process so it becomes easy and less complicated to achieve.
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APPENDIX A
MASS LOSS
The mass loss values in percentage were calculated as follows:
Mass loss percentage =
Initial mass = 10g
For table 4.2a;
550˚C
For 5 mins: mass loss= 0.17
Mass loss =1.7%
For 10 mins: mass loss= 0.25
Mass loss =2.5%
For 15 mins: mass loss= 0.42
Mass loss =4.2%
For 30 mins: mass loss= 0.82
Mass loss =8.2%
For 60 mins: mass loss= 1.02
Mass loss =10.2%
For 90 mins: mass loss= 1.10
Mass loss =11%
For table 4.2b
600˚C
For 5 mins: mass loss= 0.30
Mass loss =3%
For 10 mins: mass loss= 0.07
Mass loss =0.7%
For 15 mins: mass loss= 0.84
Mass loss =8.4%
For 30 mins: mass loss= 1.10
Mass loss =11%
For 60 mins: mass loss= 1.15
Mass loss =11.5%
For 90 mins: mass loss= 1.30
Mass loss =13%
For table 4.2c
650˚C
For 5 mins: mass loss= 0.21
Mass loss =2.1%
For 10 mins: mass loss= 0.99
Mass loss =9.9%
For 15 mins: mass loss= 1.15
Mass loss =11.5%
For 30 mins: mass loss= 0.91
Mass loss =9.1%
For 60 mins: mass loss= 1.24
Mass loss =12.4%
For 90 mins: mass loss= 1.07
Mass loss =10.7%
For table 4.2d
700˚C
For 5 mins: mass loss= 0.66
Mass loss =6.6%
For 10 mins: mass loss= 1.11
Mass loss =11.1%
For 15 mins: mass loss= 1.20
Mass loss =12%
For 30 mins: mass loss= 1.20
Mass loss =12%
For 60 mins: mass loss= 1.26
Mass loss =12.6%
For 90 mins: mass loss= 1.30
Mass loss =13%
For table 4.2e
750˚C
For 5 mins: mass loss= 0.90
Mass loss =9%
For 10 mins: mass loss= 1.22
Mass loss =12.2%
For 15 mins: mass loss= 1.27
Mass loss =12.7%
For 30 mins: mass loss= 1.24
Mass loss =12.4%
For 60 mins: mass loss= 1.28
Mass loss =12.8%
For 90 mins: mass loss= 1.30
Mass loss =13%
APPENDIX B
DEGREE OF DEHYDROXYLATION
The degree of dehydroxylation(Dtg) was calculated using the formula
Dtg =
Where
M= mass loss in %
Mmax = Loss on Ignition, LOI = 13.2%
For 550˚C
For 5mins, M=1.7%
Dtg =
For 10mins, M=2.5%
Dtg =
For 15mins, M=4.2%
Dtg =
For 30mins, M=8.2%
Dtg =
For 60mins, M=10.2%
Dtg =
For 90mins, M=11%
Dtg =
For 600˚C
For 5mins, M=3%
Dtg =
For 10mins, M=0.7%
Dtg =
For 15mins, M=8.4%
Dtg =
For 30mins, M=11%
Dtg =
For 60mins, M=11.5%
Dtg =
For 90mins, M=13%
Dtg =
For 650˚C
For 5mins, M=2.1%
Dtg =
For 10mins, M=9.9%
Dtg =
For 15mins, M=11.5%
Dtg =
For 30mins, M=9.1%
Dtg =
For 60mins, M=12.4%
Dtg =
For 90mins, M=10.7%
Dtg =
For 700˚C
For 5mins, M=6.6%
Dtg =
For 10mins, M=11.1%
Dtg =
For 15mins, M=12%
Dtg =
For 30mins, M=12%
Dtg =
For 60mins, M=12.6%
Dtg =
For 90mins, M=13%
Dtg =
For 750˚C
For 5mins, M=9%
Dtg =
For 10mins, M=12.2%
Dtg =
For 15mins, M=12.7%
Dtg =
For 30mins, M=12.4%
Dtg =
For 60mins, M=12.8%
Dtg =
For 90mins, M=13%
Dtg =
APPENDIX C
Settling time Calculation
Refining or beneficiation of the raw Okpella kaolin was aided by the use of Stoke’s law of sedimentation already discussed in the literature.
The equation is as follows:
Where ρ is the density of the particle = 2650kg/m3
ρo is the density of the solvent = 1000kg/m3
η is the viscosity of solvent at 30oC = 7.98 x 10-6Ns/m3
d is particle diameter = 20µm
h= given distance of settling tube = 34cm = 0.34m
g= gravitational acceleration = 9.81m/s2
hence, settling time, ts for quartz of 20µm
ts =
= 12minutes and 57 seconds
APPENDIX D
Formulation of Zeolite Y by the use of external silica material
The molar composition of the aluminosilicate gel used for the formulation or synthesis of zeolite Y was taken based on the work of Kovo (2010) as
15Na2O:Al2O3:15SiO2:450H2O
The mole ratio are Na2O = 15
Al2O3 = 1
SiO2 = 15
H2O = 450
From literature, metakaolin has the formula Al2Si2O7
Metakaolin has mole ratio of 1 aluminate, Al2O3 and 2 silicate, SiO2
From the molar composition of the growth gel for the synthesis of zeolite Y, the mole ratio of SiO2 is 15, hence, there arises the need to increase the SiO2 in the metakaolin up to 15, this will be done by getting the extra 15- 2= 13moles from the external silica source, say sodium metasilicate (Na2SiO3).
Metakaolin used for the process
The mass of the metakaolin is gotten as follows:
Molecular formula of metakaolin is Al2Si2O7
The relative molecular masses of individual elements are:
Al = 26.98; Si = 28.09; O = 16.00; Na= 23.00; H = 1.00
Molecular weight of metakaolin is calculated as:
= (2 x 26.98) + (2 x 28.09) +(16 x 7)
= 53.96 + 56.18 + 112
= 222.14g
mass of metakaolin = 222.14g
Amount of Additional Silica
Using sodium metasilicate (Na2SiO3), as external silica source, containing 47% SiO2 and 52% Na2O, the remaining 1% serves as impurities.
Na2SiO3 → Na2O + SiO2
Molar mass of Na2SiO3 = (2 x 23) +28.09 + (3 x 16)
= 46 + 28.09 +48
= 122.09g/mole
If 100g of Na2SiO3 is used, 47g will be SiO2
Molar mass of SiO2 = 28.09 +32 = 60.09g/mole
Amount of SiO2=
100g of Na2SiO3 = 0.78mole of SiO2
13moles will then be gotten from
= 1667.67g of Na2SiO3
Since 52% of Na2SiO3 is Na2O
Then
= 866.67g of Na2O
Converting to molar mass
Molar mass of Na2O = (23 x 2) + 16
= 62g/mole
Amount of Na2O = = 13.98 mole of Na2O
Thus remaining amount of Na2O required for the gel is 15 – 13.98 = 1.02mole of Na2O.
Additional Na2O
Additional Na2O is gotten from NaOH since Na2O does not exist naturally alone.
Thus, Na2O + H2O →2NaOH
Molar mass of NaOH = 23+16+1= 40g/mole
If 10g of NaOH is taken, then the actual amount of NaOH is 99% of 10 = 9.9g which is equivalent to = 0.248NaOH
10g = 0.248mole of NaOH
We require 1.02mole of NaOH but from the stoichiometry equation above
1mole of Na2O gives 2moles of NaOH
Hence, the required 1.02mole of Na2O will be gotten from 2 x 1.02 of NaOH = 2.04mole NaOH
10g of NaOH = 0.248mole of NaOH
xg = 2.04mole of NaOH
xg = = 82.25g of NaOH pellets
Additional H2O
1.02mole of H2O is gotten from NaOH, hence the remaining water is gotten as 450- 1.02= 448.98mole of H2O
Molar mass of H2O = 16 +2 =18g/mol
Mass of H2O required = 18 x 448.98 = 8081.64g of H2O
Thus the total amount of individual reagents required for the synthesis of zeolite Y with molar composition of 15Na2O:Al2O3:15SiO2:450H2O are as follows:
Metakaolin= 222.14g
Na2SiO3 = 1667.67g
NaOH = 82.25g
H2O = 8081.64g
Scaling down to 40g of metakaolin, we will use
Scaling factor =
Metakaolin=
Na2SiO3 = 40g
NaOH =
H2O =
= 0.1939litres
Hence the following amount of precursor were used to synthesize zeolite Y:
Metakaolin= 5.3313g
Na2SiO3 = 40g
NaOH = 1.9740g
H2O = 193.959g = 0.1939litres
APPENDIX E
Table 4.8 Degree of dehydroxylation of Okpella kaolin clay for different
calcination temperature and times
|
|
Temperatures oC
|
|
|
|
Heating time, (min)
|
|
|
|
|
|
|
550
|
600
|
650
|
700
|
750
|
5
|
0.13
|
0.23
|
0.16
|
0.50
|
0.68
|
10
|
0.19
|
0.05
|
0.75
|
0.84
|
0.92
|
15
|
0.32
|
0.64
|
0.87
|
0.91
|
0.96
|
30
|
0.62
|
0.83
|
0.69
|
0.91
|
0.94
|
60
|
0.77
|
0.87
|
0.94
|
0.95
|
0.97
|
90
|
0.83
|
0.97
|
0.81
|
0.98
|
0.98
|
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