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Serhud [2]
3 years ago
14

The production of ethylbenzene, a very popular industrial chemical, is carried out reacting benzene with ethylene in liquid phas

e. This reaction takes place in a series of reactors that involve multiple side reactions and intermediates. Ethylene, being the limiting reactant, is used up first and hence a considerable amount of benzene remains unreacted. From one of the reactors in the series, the exit stream is a mixture of this unreacted benzene (1), an intermediate – toluene (2), and the product ethyl benzene (3).
It is desirable to separate this liquid mixture before sending the components to the next series of reactors/process steps. So 100 mol/min of this mixture is flashed from 200 mm Hg and 50 °C to 100 mm Hg. If the mole fraction of benzene and toluene are 40% each when the mixture enters the flash distillation unit, determine if the mixture will flash completely, partially, or not at all. Assume ideal gas and ideal solution behavior for the vapor phase and liquid phase, respectively. If the mixture does flash partially, determine the composition and molar flow rates of the equilibrium streams exiting the reactor. Show all calculations by hand using your preferred method for solving simultaneous equations. Alternatively, you may use Solver (Excel) but this must be accompanied by a printout of a neatly formatted Excel sheet showing your equations and constraints.

Chemistry
1 answer:
gogolik [260]3 years ago
3 0

Answer:

Explanation:

CHECK BELOW ATTACHMENT FOR THE SOLUTION

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fredd [130]

Answer:

The mass of FeSO4*7H2O in the sample is 1.21 grams

Explanation:

<u>Step 1</u>: Calculate moles of Fe2O3

moles of Fe2O3 = mass of Fe2O3 / Molar mass of Fe2O3

moles of Fe2O3 = 0.348 grams / 159.69 g/mole = 0.00218 moles

<u>Step 2</u>: Calculate moles of Fe

4 Fe + 3O2 → 2Fe2O3

For 4 moles of Fe consumed there is 2 moles of Fe2O3 produced

This means it has a ratio 2:1

So 0.00218 moles of Fe2O3 produced , there is 2*0.00218 = 0.00436 moles of Fe consumed

<u>Step 3:</u> Calculate moles of FeSO4*7H2O

Fe + H2SO4 + 7H2O → FeSO4*7H20 + H2

For 1 mole of Fe consumed there is 1 mole of FeSO4*7H2O produced

This means for 0.00436 moles there is 0.00436 moles of Fe2SO4*H2O produced

<u>Step 4:</u> Calculate the mass of FeSO4*7H2O in the sample

mass of FeSO4*7H2O = 0.00436 moles * 278.01 g/mole = 1.212 g

The mass of FeSO4*7H2O in the sample is 1.21 grams

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