Bubbleplate (Hive Bee) 06-01-04 03:37 No 510691				15% NH3 in EtOH Preparation?
						At Rhodiums website, the post "Base-catalyzed Hydrolysis of Ergocristine to Lysergic Acid" (../rhodium /ergocristine.html) has a procedure for producing 15% NH3 in EtOH by adding NaOH to absolute Ethanol containing NH4Cl. However no weights of the reagents are given. If one wanted to produce say, 1 liter of 15% Ammoniated Ethanol, how much NH4Cl, NaOH, and CaO would be required?  TIA.
		Novice (Hive Bee) 06-01-04 04:29 No 510698				Try this.
						As one litre of ethanol weighs about 790 g, you want 15 percent of those grams to be ammonia, that is 790 x 0,15 = 118,5 g, which is 118,5 / 17 g/molNH3 = 7 moles roughly. NH4Cl + NaOH -> NH3 + H2O + NaCl You will need to add 7 moles of NH4Cl and 7 moles of NaOH. Hence: NH4Cl: 7 x 52,5 g/mol = 367 g. NaOH: 7 x 40,0 g/mol = 280 g. CaO: Enough to absorb 7 moles of water (CaO + H2O -> Ca(OH)2). I'm quite positive that a solution of 671,5 g ethanol and 118,5 g of ammonia does not have the volume of 1 litre, so you might have to make a slight excess just to be sure you get enough. Now that you get the idea I'm sure it won't be a problem though...
		xxxxx (Hive Bee) 06-01-04 07:50 No 510724				convienient prep
						one method is to take a 500ml jar of 28% nh4oh and a 500ml jar of c2h5oh and place them together in a 4000ml jar with a lid. after a day or two the ammonia vapors will go into the alcohol in about 15% which is about the saturation point for nh3 in c2h5oh. i got this from the tips in a lancaster catalog for preparing a silicate-free solution of nh3 from regent stored in glass container.
		Bubbleplate (Hive Bee) 06-02-04 03:11 No 510897				Thanks All!
						Appreciate the math and tips.