Nitric acid (HNO_3) is produced from ammonia, which involves the following steps

Question

Nitric acid (HNO_3) is produced from ammonia, which involves the following steps: (1) 4 NH_3 (g) + 5 O_2 (g) rightarrow 4 NO (g) + 6 H_2 O (g); (2) 2 NO (g) + O_2 (g) rightarrow 2 NO_2 (g); (3) 3 NO_2 (g) + H_2 O (l) rightarrow 2 HNO_3 (aq) + NO_(g) The NO gas produced in the last step is fed into step-2, together with the new batch of NO from step - 1 and the cycle continues. (a) Using the above equations, balance the following overall equation for the production of nitric acid from ammonia, and determine the stoichiometric mole - relationship between NH3 and HNO3. ______ NH_3 (g) + ____ O_2 (g) + H_2 O (l) rightarrow ____HNO_3 (aq) + ____NO (g) + H_2 O (g) (b) How many grams of HNO_3 are produced in one reaction cycle starting with 1.00 kg of NH_3 and excess of O_2 gas, if in one reaction cycle the yield is only 25.0%?

Explanation / Answer

common numer = 3x4 = 12 then;

12NH3 + 15O2 = 12NO + 18H2O

12NO + 6O2 = 12NO2

12NO2 + 4H2O = 8HNO3 + 4NO

add all

12NO2 + 4H2O + 12NO + 6O2 + 12NH3 + 15O2 = 12NO + 18H2O + 12NO2 + 8HNO3 + 4NO

cancel common terms

4H2O(l) + 21O2 + 12NH3 = 18H2O(g) + 8HNO3 + 4NO

b)

find mass of HNO3 if:

m = 1 kg of NH3 is reacted... yield is 25%

so

1 kg = 1000 g

mol of NH3 = mass/MW = 1000/17 = 58.82 mol of NH3

12 mol of NH3 = 8 mol of HNO3

58.82 mol --> 8/12*58.82 = 39.21 mol of HNO3

note that only 25% actually forms --> 0.25*39.21 = 9.8025 mol of HNO3

mass = mol*MW = 9.8025*63.01 = 617.65 g of HNO3

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