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.
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.
1 answer:
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Complete Question
The complete question is shown on the first uploaded image
Answer:
X is Be
Y is B
Z is O
Explanation:
In order to understand the solution, we need to that the an element form a bond based on
1 The number of valence electron available in its atom
2 The number of electron it require to attains its stable state i.e its Octet state
The number of valence electron for each of the option are
F = 7, N = 5 , O = 6 , C = 4 , B = 3 , Be = 2
Considering the first element
We see that X is donating two valence electron in its atom and it is donating it to 2 Cl as Cl only requires a single electron to achieve its stable state
Now looking option we can see that Be has two valence electron in it atom and in order for it to attain its stable state it needs to release the electron so the correct option for this question is Be
Considering the second element
We see that Y is donating three valence electron in its atom and it is donating it to 3 Cl as Cl only requires a single electron to achieve its stable state
Now looking option we can see that B has three valence electron in it atom and in order for it to attain its stable state it needs to release the valence electron so the correct option for this question is B
Considering the second element
We see that Z is donating two valence electron in its atom and it is donating it to 2 Cl as Cl only requires a single electron to achieve its stable state, also Z has 2 lone-pair valence electron
Now looking at the option we can see that O has 6 valence electron so if it donates 2 valence electron it will still be left with 4 - pair of electron
Therefore the correct option to this question is O
