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

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The discussion revolves around calculating the weight and volume of nitrogen required to produce 5.00 kg of pure nitric acid (HNO3) based on a given reaction formula. The key point is that the term "weight of nitrogen" refers to the mass of nitrogen used in the reaction. Participants clarify that to find the weight, one must calculate the number of moles of nitrogen in the nitric acid produced, noting that each mole of nitric acid is derived from half a mole of nitrogen gas (N2). The calculation yields a weight of approximately 0.556 kg of nitrogen and a volume of about 889 liters at standard temperature and pressure (STP). Additionally, it is emphasized that the reaction formula provided does not accurately represent industrial processes for producing nitric acid, which typically involve different methods.
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from a series of reactions
i got the formula which produces HNO_3... the "result" formula is:
4N_2_(_g_) + 6H_2_(_g_) + 8O_2_(_g_) -> 4HNO_3_(_a_q_) + 4H_2O_(_l_)

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 N_2 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 HNO_3... the "result" formula is:
4N_2_(_g_) + 6H_2_(_g_) + 8O_2_(_g_) -> 4HNO_3_(_a_q_) + 4H_2O_(_l_)

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 HNO_3... the "result" formula is:
4N_2_(_g_) + 6H_2_(_g_) + 8O_2_(_g_) -> 4HNO_3_(_a_q_) + 4H_2O_(_l_)

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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