Question

Nitric acid is often manufactured from the atmospheric gases nitrogen and oxygen, and hydrogen prepared by reforming natural gas, in a two-step process. In the first step, nitrogen and hydrogen react to form ammonia: N 2 ( g ) + 3 H 2 ( g ) → 2 NH 3 ( g )

In the second step, ammonia and oxygen react to form nitric acid HNO 3 and water: NH 3 ( g ) + 2 O 2 ( g ) → HNO 3 ( g ) + H 2 O ( g )

Suppose the yield of the first step is 86. % and the yield of the second step is 74. % . Calculate the mass of hydrogen required to make 8.0 kg of nitric acid. Be sure your answer has a unit symbol, if needed, and is rounded to 2 significant digits.

Answer 1

Step-1 : N₂ + 3H₂   2NH_{3 (g)} Yield = 86% and suppose both N2 and H2 are limiting reagent.

mol wt of H₂ = 2 g/mol and NH₃ = 17 g/mol, 3 moles of H₂ gives 2 moles of NH₃

that is 3*2 kg H₂ gives 2*17 kg of ammonia theoretically.

that is 6 kg H₂ gives 34 kg of ammonia theoretically.

Yield is 86% hence 34 * 86 / 100 = 29.24 kg NH₃ will produce from 6 kg H₂.

If we use that 29.24 kg NH₃ in Step-2    NH₃ + 2O₂   HNO_{3 (g)} + H₂O_(g)

mol wt of NH₃ = 17 g/mol and HNO₃ = 63 g/mol

hat is 17 kg NH₃ gives 63 kg of HNO₃ theoretically.

29.24 kg of NH₃ gives = 29.24 * 63 /17 = 108.36 kg of HNO₃ theoretically.

But yield of step-2 is 74%, hence 108.36 * 74 / 100 = 80.18 kg of HNO₃ will produce.

Hence for 80.18 kg HNO₃ requires 6 kg H₂ gives

Then for 8 kg HNO₃ requires how much kg H₂ gives

= 8 kg * 6 kg / 80.18 kg

= o.598 kg of H₂ required

= 0.60 kg of H₂ required for preparation of 8 kg HNO₃