ALVARO1

Thanks for your early reply. Yes, I understand that it is the carrier gas but what I do not understand is that it appears with three species in brackets. Why are there three compounds next to the M? Where do they come from?

Hi, I have a question and I don't know if a chemist can help me: In table 1 on page 4 (line 8), I don't understand why there are three compounds next to the M. What does the bracket mean? N2H3 + M [N2 (2.4); NH3 (3); N2H4 (4)] = N2H2 + H + M (I have attached the document) Thermal_decomposition_of_ammonia_N2H4_-_An_interme.pdf Thanks in advance, Regards.

I want to use ammonia with a concentration of 29.4% in water. I have tried to calculate it like this: Normalized stoichiometric ratio Indicates the actual amount of reagent required to achieve the NOx reduction goal. NSR is defined as: NSR= Mol of NH3 equivalent injected / Mol NOx not controlled Estimation of reagent consumption and tank size Once the NSR is estimated, the reagent consumption rate, expressed in pounds per hour (lb / hr), can be calculated using: where Mreactive is the molecular weight of the reagent ( 17.07 grams per mole of ammonia) and MNOx is the molecular weight of NO2 (46.01 grams per mole). The molecular weight of NO2 is used because emissions of NOx, NOx in, are given in lb / MMBtu of NO2. For ammonia SRT=1 For ammonia, the mass flow ratio of the aqueous reagent solution is given by: where Csol is the concentration of the aqueous reagent solution, by weight. The volumetric flow rate of the solution, generally expressed in gallons per hour (gph), is given by: Finally the volume of the tank is obtained I do not know if with these calculations it can be calculated. If someone can tell me if I'm doing well, I would appreciate it. Other problem I have is that I don't know how to calculate the heat supply "QB" from my reactor (boiler) to get the mass flow of the reactant. "QB" is the maximum potential release of the reactor (boiler) Thank you so much in advance.

Hello, I want to react NO with UREA to reduce NO whose reaction is: 2NO + CO (NH2) 2 + ½ O2--------------------->2N2 + CO2 + 2H2O The fact is that instead of UREA I have used NH3 to make it easier, then the reaction would be the following: 2NO + 2NH3 + ½ O2--------------------> 2N2 + 3H2O I know the mole fraction of NO which is 0.0003 and also that of 02 which is 0.01848 (Both in the gas phase since it is the gas produced from combustion). The question is, how much liquid NH3 do I have to use to reduce the maximum amount of NO. Thank you very much in advance.