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Flura [38]
3 years ago
14

An evaporation–crystallization process of the type described in Example 4.5-2 is used to obtain solid potassium sulfate from an

aqueous solution of this salt. The fresh feed to the process contains 19.6 wt% K2SO4. The wet filter cake consists of solid K2SO4 crystals and a 40.0 wt% K2SO4 solution, in a ratio 10 kg crystals/kg solution. The filtrate, also a 40.0% solution, is recycled to join the fresh feed. Of the water fed to the evaporator, 45.0% is evaporated. The evaporator has a maximum capacity of 175 kg water evaporated/s.Calculate the maximum production rate of solid K2SO4, the rate at which fresh feed must be supplied to achieve this production rate.

Chemistry
1 answer:
timofeeve [1]3 years ago
7 0

Answer:

feed = 220.77 kg/s; maximum production rate of solid crystal = 416 kg/s; the rate of supplying fresh feed to obtain the production rate = 1.6

Explanation:

Material or mass balance can be used to estimate the mass flow rates of all the streams in the diagram shown in the attached file.

Overall balance: M_{1} = 175 + 11M_{2}

Water: 0.804M_{1} = 175 + 0.6M_{2}

Using substitution method, we have:

M_{1} = 220.77 kg/s

M_{2}  = 4.16 kg/s

The maximum production rate of solid crystal is 10M_{2} = 10*4.16 = 416 kg/s

Around evaporator:

0.45M_{5} = 175

M_{5} = 175/0.45 = 389 kg/s

Around the mixing point:

M_{1} + M_{3} = M_{4} + M_{5}

Solid crystal: 0.196M_{1} + 0.4M_{3} = M_{4}

Using the last two equations, we can obtain:

0.804M_{1} + 0.6M_{3} = M_{5}

0.6M_{3} = 389 - 0.804*220.77 = 389 - 177.5 = 211.5

M_{3} = 211.5/0.6 = 352.5 kg/s

The rate of supplying fresh feed to obtain the production rate is:

\frac{M_{3}}{M_{1}} = 352.5/220.77 = 1.6

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Which of the following will have the slowest rate of diffusion at a given temperature?
Orlov [11]
So diffusion is inversely proportional to mass !
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How many moles are in 29.5 grams of Ax?
JulsSmile [24]

The number of moles present in 29.5 grams of argon is 0.74 mole.

The atomic mass of argon is given as;

Ar = 39.95 g/mole

The number of moles present in 29.5 grams of argon is calculated as follows;

39.95 g ------------------------------- 1 mole

29.5 g ------------------------------ ?

= \frac{29.5}{39.95} \\\\= 0.74 \ mole

Thus, the number of moles present in 29.5 grams of argon is 0.74 mole.

<em>"Your question seems to be missing the correct symbol for the element" </em>

Argon = Ar

Learn more here:brainly.com/question/4628363

6 0
2 years ago
The process in which an organic acid and an alcohol react to form an ester and water is known as esterification. Ethyl butanoate
egoroff_w [7]

Answer:

697 g

Explanation:

Ethanol (C₂H₅OH) and butanoic acid (C₃H₇COOH) react to form ethyl butanoate (C₃H₇COOC₂H₅) and water (H₂O).

C₂H₅OH + C₃H₇COOH → C₃H₇COOC₂H₅ + H₂O

The molar ratio of C₂H₅OH to C₃H₇COOC₂H₅ is 1:1. The moles of C₃H₇COOC₂H₅ produced from 6.00 moles of C₂H₅OH are:

6.00 mol C₂H₅OH × (1 mol C₃H₇COOC₂H₅/1 mol C₂H₅OH) = 6.00 mol C₃H₇COOC₂H₅

The molar mass of C₃H₇COOC₂H₅ is 116.16 g/mol. The mass corresponding to 6.00 mol is:

6.00 mol × (116.16 g/mol) = 697 g

7 0
3 years ago
2. Nitric oxide contains 46.66% nitrogen and 53.34% oxygen. Water contains 11.21% hydrogen and 88.79% oxygen. Ammonia contains 1
Mrrafil [7]

Answer:

The law of reciprocal proportions states that if two elements react individually with a given weight of a third element, the ratio of the masses with which they combine with the third element are either the same or a simple multiple of the ratio of the masses with which they combine with each other

The compounds formed includes;

1) Nitric oxide, NO

Nitrogen = 46.66% × 30.01 = 14

Oxygen = 53.34% × 30.01 = 16

2) Water, H₂O

Hydrogen = 11.21% × 18.01528 = 2

Oxygen = 88.79% × 18.01528 ≈ 16

3) Ammonia, NH₃

Hydrogen = 17.78% × 17.031 ≈ 3

Nitrogen = 82.22% × 17.031 ≈ 14

The ratio of nitrogen to oxygen in nitric oxide = 14:16 = 7:8

The ratio of nitrogen to hydrogen in ammonia = 14:3

The ratio in which hydrogen and oxygen combine with nitrogen = 3/16

The ratio of hydrogen and oxygen combine with each other in water = 2/16

Therefore, the ratio with which hydrogen and oxygen combine with nitrogen, is (2/3) times the ratio with which they combine with each other, which verifies the law of reciprocal proportions

Explanation:

4 0
3 years ago
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