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Kisachek [45]
3 years ago
11

what are the similarities and differences between distillation, azeotropic distillation, extractive distillation and liquid-liqu

id extraction.
Chemistry
1 answer:
DedPeter [7]3 years ago
6 0

Explanation:

- Simple Distillation: its a separation method that can be used when the two or more liquids in the mix have at least 50 degrees of difference between their boiling points.

-Azeotropic distillation: is a technique to break an azeotrope (constant boiling point mixtures), that can't be separated by simple distillation, by adding another component to generate a new azeotrope (between one initial component and the new one added) with lower boiling point.

-Extractive distillation: is a process to separate mixtures with close boiling points by adding a miscible, high boiling or none volatile solvent to increase the relative volatility of the liquids in the mix, this increases the separation factor. It differences from the azeotropic method because it doesn't form an azeotrope.

-Liquid-liquid extraction: is a method to separate compounds based on their relative solubilities in two different immiscible liquids.

After describing all the methods we can conclude that all of them are methods to separate substances based on their physical properties, this is their similarity. The difference between this method is the property it uses to separate (solubility in the case of extraction and boiling point in the case of destinations), the cases in which they bare used (when the liquids difference in boiling points is bigger [simple] or close [attractive and azeotropic]) and the formation of azeotropes (present in azeotropic and absent in extractive).

I hope you find this information useful and interesting! Good luck!

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A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0- L vessel at 300 K . The following equilibr
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Answer:

The concentration of N2 at the equilibrium will be 0.019 M

Explanation:

Step 1: Data given

Number of moles of NO = 0.10 mol

Number of moles of H2 = 0.050 mol

Number of moles of H2O = 0.10 mol

Volume = 1.0 L

Temperature = 300K

At equilibrium [NO]=0.062M

Step 2: The balanced equation

2NO(g) + 2H2(g) → N2(g) + 2H2O(g)

Step 3: Calculate the initial concentration

Concentration = Moles / volume

[NO] = 0.10 mol / 1L = 0.10 M

[H2] = 0.050 mol / 1L = 0.050 M

[H2O] = 0.10 mol / 1L = 0.10 M

[N2] = 0 M

Step 4: Calculate the concentration at the equilibrium

[NO] at the equilibrium is 0.062 M

This means there reacted 0.038 mol (0.038M) of NO

For 2 moles NO we need 2 moles of H2 to produce 1 mol N2 and 2 moles of H2O

This means there will also react 0.038 mol of H2

The concentration at the equilibrium is 0.050 - 0.038 = 0.012 M

There will be porduced 0.038 moles of H2O, this means the final concentration pf H2O at the equilibrium is 0.100 + 0.038 = 0.138 M

There will be produced 0.038/2 = 0.019 moles of N2

The concentration of N2 at the equilibrium will be 0.019 M

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