By using this formula (xNa2O:Al2O3:ySiO2:nH2O). For example in most of journals (approximate values - 6 g of sodium Aluminate , 22 g of Sodium silicate, 10 g of NaOH, 100 ml of DD water) were taken for synthesis,How one can synthesis the zeolite material by converting the molar ratio into weight ratio?
Hi .you should mutiply molar ratio by molar mass
3.165 mol Na2O needs 1 mol Al2O3
Na2O = 61.98 g/mol
Al2O3 = 101.96 g/mol
3.165 × 61.98 = 196.17 g
196.17 g Na2O needs 101.96 g Al2O3
The rest of the calculations are the same.
Hi .you should mutiply molar ratio by molar mass
3.165 mol Na2O needs 1 mol Al2O3
Na2O = 61.98 g/mol
Al2O3 = 101.96 g/mol
3.165 × 61.98 = 196.17 g
196.17 g Na2O needs 101.96 g Al2O3
The rest of the calculations are the same.
Thank You Mohsen Zolghadr for your kind response sir.
The information is very useful.
With Regards
Jayaprakash Madhu
Generally,
1 mol = its molecular weight
for ex: 1 mol NaOH means 40gms
1 molar NaOH solution means 40 gms of NaOH dissolved in water and make up to one litter.
3.165Na2O:Al2O3:1.926SiO2:128H2O
1 molar Na2O means 61.98 gms
3.165 Na2O = 3.165x61.98 = 196.16g Na2O
1 Al2O3 =101.96 gms
1mol SiO2 =60.08
1.926 SiO2 = 1.96 x 60.08 = 117.76gms
128 H20 = 128 X 18= 2304 gms H2O
3.165Na2O:Al2O3:1.926SiO2:128H2O
196.16 g Na2O : 101.96 g Al2O3 : 117.76g SiO2 : 2304 g H2O
Thank You Srinivasa Rao Ginjupalli sir, for the calculation.
Now I caught the better idea about the molar ratio calculation.
Srinivasa Rao Ginjupalli has nicely answered this question, starting from the concept of mole itself.
Accordingly, one "gram mole" of NaOH is equal to 40 gm of NaOH.
Similarly, one "milli mole" of NaOH is equal to 40 mg of NaOH.
The "concept of mole" and its clear understanding is simply indispensable for research in Chemistry or Chemical Engineering.
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