hi , am a chemical engineering student in nigeria. pls i need some help. i need to design a zeolite resin to treat 100,000 litres per hour of hard water. thanks a mill.
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Design
Started by , Nov 26 2005 09:05 AM
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#1
Posted 26 November 2005 - 09:05 AM
#2
Posted 30 November 2005 - 02:16 AM
i think this example will help:-
Example 1
How much zeolite A in its proton, H+, form would be needed to soften completely a cubic meter of water, if the hardness of the water was equivalent to 123 ppm of dissolved CaCO3. Assume 80% of the protons will be used in this exchange.
Solution
We have discussed the solution of this problem type in class. The formula for the proton form zeorlite A is H12[(AlO2)12(SiO2)12] .27H2O, and its formula (molecular) weight is (1+27+32+28+32)*12+27*18 = 1926 g/mol. Note the following:
The formula mass of CaCO3 = (40+12+48) = 100 g/mol
1 m3 H2O = 103 L H2O = 106 mL H2O = 106 g H2O
123 ppm CaCO3 = 123 g per 106 g of water.
1 mol CaCO3 2 mol H+ 1 mol z-A 1926 g z-A 100
123 g CaCO3 ----------- ----------- ---------- ---------- ---
100 g CaCO3 1 mol CaCO3 12 mol H+ 1 mol z-A 80
= 494 g zeolite A
That 80 % of protons of the zeolite A is used means that we require a little more zeolite A than stoichiometric quantities.
Discussion
Zeolites are aluminosilicates, and their structures consist of open frames as discussed above. Replacement of each Si atom by an Al atom in silicates results in having an extra negative charge on the frame. These charges must be balanced by trapping positive ions: H+, Na+, K+, Ca2+, Cu2+ or Mg2+. Water molecules are also trapped in the frame work of zeolites.
In this example, we assume that when we soak the zeolite in water containing Ca2+, and Mg2+ ions, these ions are more attrative to the zeolite than the small, singly charged protons. We further assumed that 80 percent of the protons in zeolite are replaced by other ions.
u can also fine similar examples form: http://www.science.u...alsilicate.html
Example 1
How much zeolite A in its proton, H+, form would be needed to soften completely a cubic meter of water, if the hardness of the water was equivalent to 123 ppm of dissolved CaCO3. Assume 80% of the protons will be used in this exchange.
Solution
We have discussed the solution of this problem type in class. The formula for the proton form zeorlite A is H12[(AlO2)12(SiO2)12] .27H2O, and its formula (molecular) weight is (1+27+32+28+32)*12+27*18 = 1926 g/mol. Note the following:
The formula mass of CaCO3 = (40+12+48) = 100 g/mol
1 m3 H2O = 103 L H2O = 106 mL H2O = 106 g H2O
123 ppm CaCO3 = 123 g per 106 g of water.
1 mol CaCO3 2 mol H+ 1 mol z-A 1926 g z-A 100
123 g CaCO3 ----------- ----------- ---------- ---------- ---
100 g CaCO3 1 mol CaCO3 12 mol H+ 1 mol z-A 80
= 494 g zeolite A
That 80 % of protons of the zeolite A is used means that we require a little more zeolite A than stoichiometric quantities.
Discussion
Zeolites are aluminosilicates, and their structures consist of open frames as discussed above. Replacement of each Si atom by an Al atom in silicates results in having an extra negative charge on the frame. These charges must be balanced by trapping positive ions: H+, Na+, K+, Ca2+, Cu2+ or Mg2+. Water molecules are also trapped in the frame work of zeolites.
In this example, we assume that when we soak the zeolite in water containing Ca2+, and Mg2+ ions, these ions are more attrative to the zeolite than the small, singly charged protons. We further assumed that 80 percent of the protons in zeolite are replaced by other ions.
u can also fine similar examples form: http://www.science.u...alsilicate.html
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