Ammonia is oxidized to nitric oxide in the following reaction: 4NH_3 + 5O_2 a 4N

Question

Ammonia is oxidized to nitric oxide in the following reaction: 4NH_3 + 5O_2 a 4NO + 6H_2O NH_2, O_3 and H_2O enter the reactor in the feed stream. The product stream leaving the reactor has a flow rate of 100 mole/s, containing 10 mol% NH_3 10 mole% O_2, 20% NO, and 60% H_2O Calculate the flow rates of each of the chemicals in the feed stream: NH_3_____moe/s; O_2 _____moles: H_2O_____mole/s Calculate the fractional conversion of: NH_3_____ Which chemical is the excess reactant?_____ What is the % excess of the excess reactant?_____ In the production of soybean oil, 100 kg/h of beans are fed to an extractor, along with 400 kg/h of hexane (H). The beans contain 25 wt% oil (O) and 75 wt% solids (S). The product stream from the extractor (stream B) is fed to a filter, in which all the solids are removed, and some of the oil, into the filter cake (stream C). The filter cake is 5 wt% oil, 95 wt% solids, and no hexane. The filtrate (Stream D) leaves the filter and enters the evaporator, where the liquid filtrate is partially vaporized. The liquid product (Stream E) is 90 wt% oil and remainder hexane. The vapor stream leaving the evaporator (Stream F) is 100% hexane, is completely condensed, and the liquid condensate is combined with the fresh feed of hexane (stream A). Completely label the diagram, before you do any calculations: assign variable names (with units) as needed. Calculate the total flow rate of each stream, in kg/h, and the composition (mass fractions).

Explanation / Answer

Q1.

inlet

NH3 =

O2 =

H2O =

NO =

change

NH3

O2

H2O

NO

outelt

NH3 = 10

O2 = 10

H2O = 60

NO = 20

since ratio is:

4 mol of NO = 5 mol of H2O

then

20 mol of NO = 20/4*5 = 25 mol of H2O were formed

We have a total of 60 mol of H2O in outlet, so 35 mol of H2O were initially present

20 mol of NO = 20/4*4 = 20 mol of NH3 were reacted

We have a total of 10 mol of NH3 in outlet, so 10+20 = 30 mol of NH3 were initially present

20 mol of NO = 20/4*5 = 25 mol of O2 were reacted

We have a total of 10 mol of NH3 in outlet, so 10+25= 35 mol of O2 were initially present

So...

a)

inlet:

NH3 = 30 mol/s

O2 = 35 mol/s

H2O = 35 mol/s

NO = 0 mol/s

b)

fracitonal conversion of NH3 = mol reacted/mol inlet = 20 / 35 = 0.57

c)

excess reactant --> must be O2, since at the end, 10 mol of NH3 and 10 mol of O2 are left, so O2 will finish first

% excess reactant = 35 / 30 = 116%

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