Manufacture of nitric acid

k3l

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

Borek

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.

Best,
Borek
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dextercioby

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


Daniel.

Borek

Now that you pointed at, it is obvious

The funny thing is I will never mix these things in Polish

Best,
Borek
--
http://www.chembuddy.com
Chemical calculators for labs and education
BATE - pH calculations, titration curves, hydrolisis

dextercioby

In physics "weight" has other sense.Chemists are making their own rules.

Daniel.

pack_rat2

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

garytse86

isnt the volume occupied by 1 mole of gas: 24dm3 at 298K, and 22.4 at 293K?

Speed

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