Chemistry Forum

Hi folks,

would like to verify the following calculation:
How much 63 % HNO₃ will the yield be from 1 700 tonnes NH₃ (for 24 hours process): НN₃ --> A --> B --> HNO₃

Synthesis equations (Ostwald process):
4NH₃ + 5O₂ --> 4NO + 6H₂O
2NO + O₂ --> N₂O₄
3N₂O₄ + 2H₂O --> 4HNO₃ + 2NO

moles NH₃:HNO₃ = 1:1
n (NH₃) = 100 mole => 6 300 tonnes HNO₃

Is this correct?

Thanks for helping,
Ned

You had forgotten the Percentage of the nitric acid.

what do you mean - could you please explain the mistake I do? Thanks, N.

1 700 t NH₃ / 17 = 100 mole NH₃
m(HNO₃) = 100 mole * 63 = 6 300 t HNO₃

6300 t is 100% HNO₃. But the acid is only 63%. w.w. So how much you will get.

6 300 т --> 100 % acid
x t --> 63 % => 3 969 t 63 % HNO₃

correct?

No the 63% is not the yield its the concentration. The amount must be higher. Its the opposit 10000 t 63% acid contain 6300 t 100%.

Oh, of course, the diluted acid will be more... My silly mistake Thanks a lot!