Question 1 2C2 H4 + 02 -- > 2C2 H40 Ethylene oxide is produced by the catalytic

Question

Question 1

2C2 H4 + 02 -- > 2C2 H40

Ethylene oxide is produced by the catalytic oxidation of ethylene. The feed into the reactor contains 1700 g/min ethylene and 3700 moles/min air.

a.      What is the limiting reactant?

b.      What is the extent of reaction if only 81% of the limiting reactant is converted to the product?

c.       What is the molar composition of the product stream?

Question 2

Air at 95 °F can hold maximum of 3.52 wt% water, defined as 100% humidity at 95°F. On a hot humid day the temperature is 95°F and the air contains 3.2 wt% water (90.9 % humidity). The air cooled to 68 °F because air at 68 °F can hold a maxium of 1.44 wt% water ( i.e 100% humidity at 68 °F), water condenses from the air. Hot humid air (95°F, 90.9% humidity) flows into a cooler at a rate of 154 .0 Kg/min. Calculate the flow rate of 2 streams leaving the cooler:

a.      Air at 68°F and 1OO% humidity (stream 1: air leaving the cooling unit)

b.      Water condensed from the air (stream 2: water leaving the cooling unit)

c.       What is the degree of freedom for the cooling unit

Question 3

A mixture of organic solvents containing 45.0 mole o/o xylene, 25.0o/o toluene, and the balance benzene (X) is fed to a distillation column. The bottom product contains 98.0 mole% xylene and no benzene, and 96.0°/o of the xylene in the feed is recovered in this stream. The overhead product is fed to a second distillation column. The overhead product from the second column contains 97.0o/o of the benzene in the feed to this column. The composition of this stream i 94.0 mole% benzene and the balance toulene.

a)      Draw and label a flow chart for the process

b)      Calculate the unknown process variables

c)      The percentage of the benzene in the process feed (i.e. the feed to the first column) that emerge ,in the overhead product from the second column.

d)      The percentage of toluene in the process feed that emerges in the bottom product from the second column.

If anybody can solve these questions I would rally appreciate it.

Explanation / Answer

There are multiple questions here . i am allowed to answer only 1 at a time. I will answer 1st question for you.Please ask other as different question

1.
a)
In 1 min:
Number of moles of O2 = 3700 moles
number of moles of ethylene = mass of ethylene /molar mass of ethylene
                                                            = 1700 / (28)
                                                             = 60.7 mol

Clearly ethylene is limiting reagent

b)
81% of ethylene reacts.

So extent of reaction will be 81%

c)
81% of ethylene reacts.
Moles of ethylene reacted = 0.81*60.7 = 49.2 mol
Moles of C2H4O formed = 49.2 mol

Moles of air reacted = 0.5*Moles of ethylene reacted since 1 mol of air needs 2 mol of ethylene
                                           = 0.5*49.2
                                            =24.6 mol

So product contains:
Moles of ethylene = 60.7 - 49.2 = 11.5 mol/min
Moles of air = 3700 - 24.6 =3675.4 mol/min
Moles of C2H4O = 49.2 mol/min

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