The three reactions involved in the Ostwald process for the conversion of NH3 to HNO3 are:
4NH3(g) + 5O2(g) ==> 4NO(g) + 6H2O(l) ------(1)
2NO(g) + O2(g) ==> 2NO2(g) ------------------------(2)
3NO2(g) + H2O(l) ==> 2HNO3(aq) + NO(g) ---(3)
Mass of HNO3 produced = 1.81 tons
In grams, the mass of HNO3 = 1.81*2000*453.6 = 1642032 g
Molar mass of HNO3 = 63 g/mol
Thus, # moles of HNO3 produced = 1642032/63 = 26064 moles
Based on the stoichiometry of reaction (3):
Theoretical moles of NO2 = 1.5(Moles HNO2 ) = 1.5(26064) = 39096
Assuming 80% yield:
Actual moles of NO2 = 0.8*39096 = 31277 moles
Based on the stoichiometry in reaction (2):
Theoretical moles of NO = 31277 moles
Assuming 80% yield:
Actual moles of NO = 0.8*31277 = 25022 moles
Bases on stoichiometry in reaction(3):
Moles of NH3 = 25022 moles
Assuming 80% yield:
Actual moles of NH3 required = 0.8*25022 = 20018 moles
Mass of NH3 required = 20018 moles * 17 g/mole = 340306 g = 340.31 kg
