Asked by Susan
Nitric acid can be produced from NH3 in three steps process:
I)4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
II)2NO(g) + O2(g) → 2NO2(g)
III)3NO2(g) + H2O(l) →2HNO3(aq) + NO(g)
% yield of Ist, IInd & IIIrd are respectively 50 % , 60 % & 80 % respectively the what volume of NH3(g) at STP required to produced 1575 kg of HNO3.
I)4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
II)2NO(g) + O2(g) → 2NO2(g)
III)3NO2(g) + H2O(l) →2HNO3(aq) + NO(g)
% yield of Ist, IInd & IIIrd are respectively 50 % , 60 % & 80 % respectively the what volume of NH3(g) at STP required to produced 1575 kg of HNO3.
Answers
Answered by
DrBob222
1575 kg
1575000g
1575000/17 = estimated 9E4 mol
All of the NH3 at the beginning produces HNO3 (at 100% yield) so 9E4 mol NH3 will produce 9E4 mol HNO3. That will be 9E4 mols x 22.4 mol/L = estd 2E6 L if everything is 100%. But these are not. The efficiency is 0.5 x 0.6 x 0.8 = 0.24 so
2E6L at 100%/0.24 = ? volume needed at 24%. Note: That 2E6 L is an estimate. You need to sharpen that number.
1575000g
1575000/17 = estimated 9E4 mol
All of the NH3 at the beginning produces HNO3 (at 100% yield) so 9E4 mol NH3 will produce 9E4 mol HNO3. That will be 9E4 mols x 22.4 mol/L = estd 2E6 L if everything is 100%. But these are not. The efficiency is 0.5 x 0.6 x 0.8 = 0.24 so
2E6L at 100%/0.24 = ? volume needed at 24%. Note: That 2E6 L is an estimate. You need to sharpen that number.
Answered by
Indra
3500 Litre
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