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Monica [59]
3 years ago
8

We get 1700 Tonnes of Ammonia every day. How many tonnes of 63% Nitric acid we can get?

Chemistry
1 answer:
jeyben [28]3 years ago
8 0
There are 3 equations involved in manufacturing Nitric Acid from Ammonia. 

First the ammonia is oxidized:
4NH3 + 5O2 = 4NO + 6H2O

Then for the absorption of the nitrogen oxides.
2NO + O2 = N2O4

Lastly, the N2O4 is further oxidized into Nitric acid.
3N2O4 + 2H2O = 4HNO3 + 2NO

Then run stoichiometry through these equations.
The first equation produces roughly 271,722,938 grams of NO
The second equation produces roughly 416,606,944 grams of N2O4
The last equation produces roughly 380,412,294 grams of HNO3 (nitric acid)

Convert the exact number back into tons, and your answer is: <span>419.332775 tons.
</span>
Rounded, I'm going to say that's 419.33 tons.
Hope this helps! :)

Also, it seems that commercially, Nitric Acid is commonly made by bubbling NO2 into water, rather than using ammonia.
 
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0.045 moles of NaOH

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This kind of concentration shows the relation between moles of solute which are contained in 1L of solution.

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(ii) 1 dye

(iii) Food coloring F is insoluble in the solvent

(iv) 'E' and 'H'

(b) Food colouring G

Explanation:

Paper chromatography principle is based on the rates of migration of chemicals across a sheet of paper which are different and it consists of a stationary phase such as the water in the paper and a mobile phase such as the solvent resulting in the partitioning of the components of the mixture across the paper

The solution components are positioned to start in one place from where they migrate and separate out on the chromatography paper

(ii) The number of components into which the food colouring 'H' separates into = 1

Therefore, the number of dyes in food colouring 'H' = 1 dye

(iii) Food coloring 'F' does not move because it is insoluble in the solvent, which is the mobile phase

(iv) The food colouring that contains the dye that is likely to be most soluble in the solvent are does for which the dyes travel furthest, which are;

Food coloring 'E' and 'H'

(b) Using a similar question solution found on 'tutor my self' website, we have;

The R_f values are given as follows;

R_f = \dfrac{Distance \ moved \ by \ dye}{Distance \ moved \ by \ solvent}

The distance moved by the solvent = 5 units

The distance moved by dyes in food colouring 'E' and 'H' = 4 units

The distance moved by dye in food colouring 'G' = 3.3 units

The distance moved by the second dye in food colouring 'E' = 2.7 units

By inspection, we get;

R_f dye in food colouring 'G' = 3.3/5 = 0.66,

Therefore, the dye with R_f value closest to 0.67 is the dye in food colouring 'G'.

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