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.