Eleven hundred kg/h of an ethanol/methanol/propanol stream are fed to a distillation column. twenty percent of the feed is ethanol. ninety percent of the methanol is to be recovered in the distillate along with 50% of the ethanol. all of the 400 kg/h of propanol fed to the process must be sent to the bottom. calculate all unknown stream flow rates and composition.

Feed steam is 1100 kg/h

20% of Ethanol (220 kg/h)  

36.36% of propanol (400 Kg/h)

43.64% of methanol (480 Kg/h)

Distilled steam is 542 Kg/h

20.30 % of etanol (110 kg/h)  

79.70 % of methanol (432 kg/h)  

Bottom steam is 558 Kg/h

19.71% of Ethanol (110 kg/h)  

8.60% of propanol (400Kg/h)

71.68% of methanol (48 Kg/h)

Explanation:

The attached figure shows the distillation scheme representation. There are three steams, feed, distillate and Bottom, the feed and Bottom steam have ethanol, methanol and propanol, and the distillate steam only has ethanol and methanol.

Solving Feed Steam:  

The problem says that 20% of the feed is ethanol:

Ethanol feed= (1100 Kg/h)*(0.20)=  220 Kg/h  

The problema also says that propanol feed is 400 kg/h, so the sum of ethanol+methanol+propanol=1100  

Methanol= (1100- 400-220)kg/h= 480 Kg/h  

Solving Distillate steam

The problem says that 90% of the feed methanol was recovered in distillate steam  whit the 50% of the feed ethanol  

Methanol= (480 Kg/h)*(0.9)=432 Kg/h

Ethanol= (220 Kg/h)*(0.5)= 110 Kg/h

Solving Bottom steam  

If the 90% of the feed methanol and the 50% of the feed ethanol were recovered in distillate steam, so 10% and 50% were recovered in Bottom steam, respectively. Also, the problem says all the propanol was recovered in Bottom steam

Methanol= (480 Kg/h)*(0.1)=48 Kg/h

Ethanol= (220 Kg/h)*(0.5)= 110 Kg/h

Propanol= 400 Kg/h

Answer:

Step-by-step explanation:

It is clear by the equation 2(27+3×35.5)= 267 gm of AlCl3 reacts with 6× 80 = 480 gm of Br2 . So 29.2 gm reacts = 480× 29.2/267= 52.6 gm

glycoproteins

Explanation:

A positive reaction for Molisch's test is given by almost all carbohydrates (exceptions include tetroses & trioses). It can be noted that even some glycoproteins and nucleic acids give positive results for this test (since they tend to undergo hydrolysis when exposed to strong mineral acids and form monosaccharides).

Answer:

Taking into accoun the ideal gas law, The volume of a container that contains 24.0 grams of N2 gas at 328K and 0.884 atm is 26.07 L.

An ideal gas is a theoretical gas that is considered to be composed of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P×V = n×R×T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas. The universal constant of ideal gases R has the same value for all gaseous substances.

Explanation:

In this case, you know:

P= 0.884 atm

V= ?

n= 0.857 moles (where 28 g/mole is the molar mass of N₂, that is, the amount of mass that the substance contains in one mole.)

R=0.082

T= 328 K

Replacing in the ideal gas law:

0.884 atm×V= 0.857 moles× 0.082 ×328 K

Solving:

V= 26.07 L

The volume of a container that contains 24.0 grams of N2 gas at 328K and 0.884 atm is 26.07 L.