A proposed process to remove H2S is by reaction with SO2: 2 H2S + SO2 ® 3 S + 2 H2O
In a test of the process, a gas stream containing 20% H2S and 80% CH4 was combined with a stream of pure SO2. The process produced 5000 lb of S, and in the product gas the ratio of SO2 to H2S was equal to 3, and the ratio of H2O to H2S was 10. You are asked to determine the fractional conversion of the limiting reactant, and the feed rates of the H2S and SO2 streams.
Basis: 5000 lb S = (5000 lb) / (32 lb/lb mol) = 156.25 lb mol 2 H2S + SO2 ® 3 S + 2 H2O input 114.587 83.331
react 104.17 52.08 156.25 104.17 output 10.417 31.251 156.25 104.17
conversion of the limiting reactant (H2S) =
= (104.17/114.587) x 100% = 91%
the feed rates of the H2S and SO2 streams = 114.587 and 83.441 mol REACTOR
SO2
CH4 80%
H2S 20%
S
CH4
H2S S H2O
H2O/ H2S = 10 104.17/ H2S = 10 H2S = 10.417 SO2/ H2S = 3
SO2/10.417= 3 SO2 = 31.251
mol of H2S that react
mol of H2S introduced X 100%
Formaldehyde (CH2O) is produced industrially by the catalytic oxidation of methanol (CH3OH) according to the following reaction:
CH3OH + ½ O2® CH2O + H2O
Unfortunately, under the conditions used to produce formaldehyde at a profitable rate, a significant portion of the formaldehyde reacts with oxygen to produce CO and H2O, that is:
CH2O + ½ O2 ® CO + H2O
Assume that methanol and twice the stoichiometric amount of air needed for complete conversion of the CH3OH to the desired products (CH2O and H2O) are fed to the reactor. Also assume that 90% conversion of the methanol results, and that a 75% yield of formaldehyde occurs based on the theoretical production of CH2O by reaction 1. Determine the composition of the product gas leaving the reactor.
Basis: 1 mol CH3OH
CH3OH + ½ O2 ® CH2O + H2O input 1 0.5 1
react (0.9)(1)=0.9 0.45 0.9 0.9
output 0.1 0.55 0.9 0.9
CH2O + ½ O2 ® CO + H2O
input 0.9 0.55
react 0.9-0.75=0.15 0.075 0.15 0.15
output 0.75 0.475 0.15 0.15
yield = 75% = (Z/1) x 100%, Z = 0.75 Composition of product
CH3OH = 0.1 O2 = 0.475 CH2O = 0.75 CO = 0.15
H2O = 0.9 + 0.15 = 1.05
N2 = (79/21) x 1 mol O2 in feed = 3.76 Total = 6.285
limiting reactant
Composition in % mol
CH3OH = (0.1/6.285) x 100% = 1.59%
O2 = (0.475/6.285) x 100% = 7.56%
CH2O = (0.75/6.285) x 100% = 11.93%
CO = (0.15/6.285) x 100% = 2.39%
H2O = (1.05/6.285) x 100% = 16.71%
N2 = (3.76/6.285) x 100% = 59.82%
A reactor is used to oxidize SO2 to SO3. 20% redundant feed air and only 90% SO2 conversion. Determine the composition of the gas produced by the reactor.
Basis: 100 mol F1
SO2 + ½ O2 ® SO3
input 100 !"#!###,%! (100) = 60
reacts !##&# (100) = 90 45 90
output 10 15 90
Composition gas product SO2 = 10
SO3 = 90 O2 = 15
N2 = (79/21)(60) = 225.71 Total = 340.71
reactor
F1 = mol
SO2
excess air: 20%
N2 79%
O2 21%
P = mol SO2 SO3
O2 N2
Composition in % mol
SO2 = (10/340.71) x 100% = 2.93%
SO3 = (90/340.71) x 100% = 26.42%
O2 = (15/340.71x 100% = 4.40%
N2 = (225.71/340.71) x 100% = 66.25%
Acrylonitrile is produced by reacting propylene, ammonia, and oxygen, according to the following reaction:
C3H6 + NH3 + 1,5 O2®C3H3N + 3 H2O
The reactor feed contains 10% propylene, 12% ammonia, and 78% air.
Determine:
a. limiting reactant b. % excess reactant
c. mole ratio of acrylonitrile/NH3 feed if the conversion is 30%
d. composition of product leaving the reactor
Basis: 100 mol F
C3H6 + NH3 + 1½ O2 ® C3H3N + 3 H2O
input 10 12 16.38
reacts 10 10 15 10 30
output 10 2 1.38 10 30
limiting reactant = C3H6
excess reactant = NH3 entering–NH3 required NH3 required
= !"#!$
!$ x 100% = 20%
O2 entering –O2 required O2 required
= !%,'(#!)!) x 100% = 9.2%
F= mol C3H6 10%
NH3 12%
Air: 78%
N2 (79%)(78%) = 61,62%
O2 (21%)(78%) = 16,38%
P = mol C3H6
NH3 O2
N2
C3H3N H2O reactor
X 100%
X 100%
If conversion 30% =
C3H6 + NH3 + 1½ O2 ® C3H3N + 3 H2O
input 10 12 16.38
reacts (0,3)(10) = 3 3 4.5 3 9
output 7 9 11.88 3 9
mole ratio of acrylonitrile/NH3 feed = 3/10 composition of product leaving the reactor : Composition gas product
C3H6 = 7 NH3 = 9 O2 = 11.88 C3H3N = 3 H2O = 9 N2 = 61.62 Total = 101.5
Example material balances involving combustion Coal has a composition as follows (% mol)
C = 65.4 % H = 5.3 % O = 18.5 % N = 10.2 % S = 0.6 %
is combusted with dry air. All S burns to SO2. Composition of dry gas produced :
CO2 + SO3 = 13.00 %
CO = 0.76 %
O2 = 6.17 %
H2 = 0.87 %
N2 = 79.20 %
Determine :
a. lb coal combusted/100 lb mol dry gas produced b. lb H2O generated/100 lb coal combusted c. % excess air
Composition in % mol
C3H6 = (7/101.5) x 100% = 6.89%
NH3 = (9/101.5) x 100% = 8.87%
O2 = (11.88/101.5) x 100% = 11.70%
C3H3N = (3/101.5) x 100% = 60.71%
H2O = (9/101.5) x 100% = 2.96%
N2 = (61.62/101.5) x 100% = 8.87%
