How Much Nitrogen is Required to Produce 5kg of Nitric Acid?

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Discussion Overview

The discussion centers around the calculation of nitrogen required to produce 5 kg of nitric acid (HNO3), exploring the interpretation of "weight of nitrogen" in the context of chemical reactions and stoichiometry. Participants examine the chemical formula involved and the implications of gas volumes at standard temperature and pressure (STP).

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant presents a reaction formula for the production of nitric acid and seeks clarification on the meaning of "weight of nitrogen."
  • Another participant interprets "weight" as the literal mass of nitrogen used, suggesting to treat it as if it were a solid.
  • A different participant emphasizes that in chemistry, "weight" and "mass" are not strictly synonymous and cautions against taking the term literally.
  • One participant challenges the relevance of the provided reaction formula, stating that industrial production of nitric acid involves different processes and that the calculation should focus on moles of nitrogen needed for the given mass of nitric acid.
  • There is a discussion about the volume occupied by gases at different temperatures, with a participant questioning the standard volume values.
  • A later reply asserts that the only source of nitrogen in nitric acid is atmospheric nitrogen (N2) and provides specific calculations for the weight and volume of nitrogen needed.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of "weight of nitrogen" and the relevance of the reaction formula. There is no consensus on the best approach to the problem, and multiple competing views remain throughout the discussion.

Contextual Notes

Some participants highlight the dependence on definitions of weight and mass, and there are unresolved questions regarding the accuracy of gas volume measurements at different temperatures.

k3l
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from a series of reactions
i got the formula which produces [tex]HNO_3[/tex]... the "result" formula is:
[tex]4N_2_(_g_) + 6H_2_(_g_) + 8O_2_(_g_) -> 4HNO_3_(_a_q_) + 4H_2O_(_l_)[/tex]

The question asks to find the weight of nitrogen,
and the Volume of nitrogen (at STP) required to produce 5.00 kg pure nitric acid.

what i want to know is what does it mean by the weight of nitrogen?

thx
 
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k3l said:
what i want to know is what does it mean by the weight of nitrogen?

Literally - weight of the nitrogen used. I suppose if it was a solid you will have no problems? So do it exactly the same way as if [tex]N_2[/tex] was solid.

And don't forget that 1 mole of gas at STP has volume 22.4 l.Borek
 
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"Weight+chemistry=mass".Don't take it literally or as something axiomatic...


Daniel.
 
dextercioby said:
"Weight+chemistry=mass". Don't take it literally or as something axiomatic...

Now that you pointed at, it is obvious :blushing:

The funny thing is I will never mix these things in Polish :smile: Borek
 
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In physics "weight" has other sense.Chemists are making their own rules.

Daniel.
 
k3l said:
from a series of reactions
i got the formula which produces [tex]HNO_3[/tex]... the "result" formula is:
[tex]4N_2_(_g_) + 6H_2_(_g_) + 8O_2_(_g_) -> 4HNO_3_(_a_q_) + 4H_2O_(_l_)[/tex]

The question asks to find the weight of nitrogen,
and the Volume of nitrogen (at STP) required to produce 5.00 kg pure nitric acid.

what i want to know is what does it mean by the weight of nitrogen?

thx
That "formula" may describe the overall ratio of the elements involved, but it's NOT how HNO3 is made. Industrially, it's made by the catalytic oxidation of N2 to nitrogen oxides, which are dissolved in H2O to produce a mixture of HNO3/HNO2. The HNO2 is then oxidized to HNO3 with O2 from air. In the lab, HNO3 is made by heating a nitrate with H2SO4 and distilling off the vapor. But, you DON'T NEED the reaction formula to answer that question! You just need to calculate the number of moles of N in 5 kg of HNO3. Half that number will be the moles of N2 needed, and the volume is found by V = n*R*T/P
 
isnt the volume occupied by 1 mole of gas: 24dm3 at 298K, and 22.4 at 293K?
 
k3l said:
from a series of reactions
i got the formula which produces [tex]HNO_3[/tex]... the "result" formula is:
[tex]4N_2_(_g_) + 6H_2_(_g_) + 8O_2_(_g_) -> 4HNO_3_(_a_q_) + 4H_2O_(_l_)[/tex]

The question asks to find the weight of nitrogen,
and the Volume of nitrogen (at STP) required to produce 5.00 kg pure nitric acid.

what i want to know is what does it mean by the weight of nitrogen?

thx

Listen to pack_rat2. You don't need the overall reaction because the only source of nitrogen in nitric acid is atmospheric nitrogen, or N2. 1 mole of nitric acid is made by half a mole of nitrogen.

Weight of nitrogen = 0.556 kg
Volume of nitrogen = 889 litres
 
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